Practice Test 2
Choose the best answer
- State Newton's second law.
- An object will remain at rest or move in a
straight line with a costant speed unless
acted upon by a net force.
- Any two objects in the universe attract each
other with a force directly proportional to
the product of their masses and inversely
proportional to the square of the distance
separating them.
- The force acting on an object is inversely
proportional to the acceleration produced by
the force.
- The force acting on an object is directly
proportional to the acceleration produced by
the force.
- For every reaction, there is an equal but
opposite reaction.
Answer: D
- Calculate the force acting on a 6.443 kg object
moving with an acceleration of 24 m ⁄ s 2.
- 126.822 N
- 69.669 N
- 229.93 N
- 154.632 N
- 258.491 N
Answer: D
- A force that causes an acceleration of 7 m ⁄ s 2
on an object of mass 20.123 kg is acting on a
50 kg object. Calculate the acceleration of the 50 kg object.
- 4.167 m ⁄ s 2
- 2.817 m ⁄ s 2
- 5.069 m ⁄ s 2
- 2.412 m ⁄ s 2
- 3.399 m ⁄ s 2
Answer: B
- The component of the surface force between two
surfaces sliding on each other parallel to
the surface is called
- friction
- gravitational force
- normal force
- contact force
- sliding force
Answer: A
- An object of mass 48.678 kg is sliding on a horizontal
surface with a uniform speed. The coefficient
of kinetic fiction of the surfaces is 0.2. Calculate
the force of friction exerted by the surface on the object.
- 173.387 N
- 42.622 N
- 95.409 N
- 137.284 N
- 70.569 N
Answer: C
- An object of mass 18.333 kg is being pulled by a
horizontal force of 210 N on a horizontal force.
The coefficient of kinetic friction between
the object and the surface is 0.45. Calculate the
acceleration of the object.
- 7.045 m ⁄ s 2
- 8.396 m ⁄ s 2
- 9.969 m ⁄ s 2
- 11.447 m ⁄ s 2
- 5.149 m ⁄ s 2
Answer: A
- An object is being pulled on a horizontal
surface by a horizontal force of 300
N.
If it s moving with a uniform speed, calculate the
force of friction exerted by the surface on the object.
- 30.612 N
- 300 N
- 3234 N
- 360 N
- 2940 N
Answer: B
- An object of mass 280 kg is being pulled on a friction less
horizontal surface by means of two strings. One of the strings
is pulling forward, has a tension of 600 N and makes
an angle of 30 degree with the horizontal-right. The other string is
pulling backwards horizontally and has a tension of 20
N.
Calculate the acceleration of the object.
- 2.05 m ⁄ s 2
- 1.784 m ⁄ s 2
- 3.218 m ⁄ s 2
- 1.176 m ⁄ s 2
- 2.351 m ⁄ s 2
Answer: B
- An object of mass 8 kg is being pulled on a horizontal
surface by a string that makes an angle of
40 degree with the horizontal. The tension in the string is 17 N.
Calculate the normal force exerted by the ground on the object.
- 67.473 N
- 49.729 N
- 59.987 N
- 27.622 N
- 122.067 N
Answer: A
- A 16 kg object and a 7 kg object are
attached by a string. If the 7 kg object is being pulled
by a force of 220 N. Calculate their acceleration.
- 9.565 m ⁄ s 2
- 0.976 m ⁄ s 2
- 7.652 m ⁄ s 2
- 31.429 m ⁄ s 2
- 13.75 m ⁄ s 2
Answer: A
- The SI unit of measurement for work is
- Volt
- Coloumb
- Joule
- Pascal
- Newton
Answer: C
- An object is being pulled to the left by a 80 N force
that makes an angle of 70 deg with the horizontal-left
while moving to the right for a distance of 50 m.
Calculate the work done by the force.
- -1368.081 J
- -1094.464 J
- 1094.464 J
- 1368.081 J
- -1641.697 J
Answer: A
- Calculate the speed of an object of mass 42.333 kg
if its kinetic energy is 1800 J.
- 4.521 m ⁄ s
- 1.442 m ⁄ s
- 13.754 m ⁄ s
- 9.222 m ⁄ s
- 5.523 m ⁄ s
Answer: D
- Under the influence of some forces, the speed
of a 21.325 kg object changed from 6 m ⁄ s to a certain speed.
If the net work done on the object is 1450 J,
Calculate its final speed.
- 13.115 m ⁄ s
- 15.634 m ⁄ s
- 19.322 m ⁄ s
- 7.263 m ⁄ s
- 21.714 m ⁄ s
Answer: A
- An object has a kinetic energy of 120 J as it crosses
a location where its potential energy is 19 J.
Calculate its mechanical energy as it crosses this location.
- 83.4 J
- 101 J
- 139 J
- 111.2 J
- 111.1 J
Answer: C
- Under the influence of conservative forces only,
an object is displaced from point A to point B.
Its potential energy and kinetc energy at point A are
respectively 52.666 J and 60.554 J. Its kinetic energy at point B is
40 J. Calculate its potential energy at point B.
- 16.563 J
- 114.739 J
- 133.294 J
- 73.22 J
- 85.14 J
Answer: D
- An object of mass 12.333 kg is located at the top
of a 24 m tall building. Calculate the
gravitational potential energy of the object
with respect to a point 1 m below the top
of the building .
- 70.278 J
- 120.863 J
- 222.484 J
- 186.338 J
- 45.915 J
Answer: B
- A roller coaster extends to the ground
from a height of 40 m (point A) and then rises
to a height of 19 m (point B). An object of mass 2 kg
starts at point A with a speed of 8 m ⁄ s. Assuming
the roller coaster is friction less, calculate the
speed of the object by the time it reaches point B.
- 5.494 m ⁄ s
- 9.091 m ⁄ s
- 2.238 m ⁄ s
- 21.808 m ⁄ s
- 38.062 m ⁄ s
Answer: D
- A spring extends by 0.8 m when an object of mass
3 kg hangs from it. By how much will it extend
when an object of mass 14 kg hangs from it.
- 3.001 m
- 3.733 m
- 1.819 m
- 5.291 m
- 6.321 m
Answer: B
- Work done by the non conservative forces acting
on an object is equal
- to the change in the mechanical energy of the object
- to the change in the kinetic energy of the object
- to the work done by the conservative forces
- to the change in the potential energy of the object
- to the net work done on the object
Answer: A
- Intially an object has a potential energy of 80 J
and a kinetic energy of 190 J. Under the influence
of conservative and non-conservative forces
its potential energy changed to 30 J and its kinetic
energy changed to 36 J. Calculate the work done
by the non-conservative forces.
- -196 J
- -209 J
- -204 J
- -201 J
- -198 J
Answer: C
- Calculate the momentum of an object of mass 0.36 kg
moving with a speed of 4.8 m ⁄ s.
- 1.728 kg m ⁄ s
- 0.488 kg m ⁄ s
- 1.519 kg m ⁄ s
- 2.792 kg m ⁄ s
- 1.15 kg m ⁄ s
Answer: A
- An object of mass 0.16 kg falling vertically downward
hits the ground with a speed of 12 m ⁄ s and bounces
back vertically upward with a speed 3 m ⁄ s. If the object was in
contact with the ground for 0.2 seconds, calculate the
average force exerted by the ground on the object.
- 9.933 N
- 21.839 N
- 12 N
- 7.18 N
- 16.084 N
Answer: C
- An object of mass 9.5 kg was acted by a force of 24.2 N for
0.6 seconds. If its initial speed was 2.0 m ⁄ s,
calculate its final speed.
- 1.745 m ⁄ s
- 3.528 m ⁄ s
- 1.101 m ⁄ s
- 4.241 m ⁄ s
- 6.311 m ⁄ s
Answer: B
- An object of mass 4 kg moving with a speed
of 25 m ⁄ s to the right collides with an object
of mass 17 kg moving with a speed of 12 m ⁄ s
to the left. After collision, the 17 kg object moves
to the right with a speed of 9 m ⁄ s to the right. Calculate
the velocity of the 4 kg object after collision.
- -121.326 m ⁄ s
- -71.691 m ⁄ s
- -103.327 m ⁄ s
- -92.482 m ⁄ s
- -64.25 m ⁄ s
Answer: E
- An object of mass 16 kg moving with a speed
of 28 m ⁄ s to the right collides with an object
of mass 17 kg at rest. If the collision is completely inelastic,
calculate the kinetic energy lost during the collision.
- -3231.03 J
- -1009.728 J
- -5972.282 J
- -4411.118 J
- -1871.373 J
Answer: A
- Which of the following statements is a correct
statement.
- Colliding objects always lose kinetic energy
during collision.
- All collisions where momentum is conserved are
called completely elastic collisions.
- If colliding objects are treated as a single
system, then the net force acting on the
system is zero.
- A collision where both momentum and kinetic
energy are conserved is called a completely
inelastic collision.
- Kinetic energy is always conserved during a
collision.
Answer: C
- After a bullet of mass 0.06 kg is fired into a
ballistic pendulum of mass 1.5 kg, the bullet is embedded
in the pendulum and the pendulum rose to a height of 0.25 m.
Calulate the speed with which the bullet was fired into the
ballistic pendulum.
- 57.553 m ⁄ s
- 93.151 m ⁄ s
- 108.616 m ⁄ s
- 41.241 m ⁄ s
- 49.171 m ⁄ s
Answer: A
- An object of mass 11 kg going to the right with
a speed of 30 m ⁄ s collides with a(n) 16 kg object going to
the left with a speed of 5 m ⁄ s. If the collision
is completely elastic, calculate the speed of the 16 kg
object after collision.
- 15.128 m ⁄ s
- 10.224 m ⁄ s
- 21.12 m ⁄ s
- 23.519 m ⁄ s
- 35.135 m ⁄ s
Answer: D
- An object of mass 21 kg going to the right with
a speed of 21 m ⁄ s collides with a(n) 10 kg object at rest.
After collision the 21 kg object moves with a speed of 10 m ⁄ s
making an angle of 10 degree with the horizontal-right.
Calculate the x-component of the velocity of the 10 kg
object after collision.
- 38.542 m ⁄ s
- 23.419 m ⁄ s
- 18.904 m ⁄ s
- 30.257 m ⁄ s
- 12.274 m ⁄ s
Answer: B