O lvl Phy: Work Energy Power

Catholic High Prelims 2004 P2 Q4

A toy car is given a brief push which sends it along a horizontal section of a runway. The car moves down the ramp as shown in the figure below.

(a) Draw labelled arrows on the diagram to show the 3 forces acting on the car at A.

(b) The length of the ramp is 0.90 m and the average force of friction is 0.11 N. The car has 0.30 J of kinetic energy at the top of the ramp and loses 0.50 J of potential energy as it moves from the top to the bottom of the ramp.
(i) Calculate the work done by the car against friction as it moves down the ramp.
(ii) Calculate the kinetic energy of the car at the bottom of the ramp.

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Answer:

(a)

(b)
(i)
Work done = f * d
= 0.11 * 0.90 J
= 0.099 J (2 s.f.)
= 9.9 * 10^-2 J

(ii)
Kinetic energy of the car = initial KE + loss in GPE - energy loss due to frictional forces
= 0.30 J + 0.50 J - 0.099J
= 0.70 J (2 s.f.)

O lvl Phy: Work Energy Power

RI Sec 3 EOY 1998 P2 Q4

A student releases a trolley of mass 0.20 kg from the top of an inclined runway of height 0.40 m as shown in the figure below.

The speed of the trolley at the bottom of the runway is 1.0 m/s

(a) Calculate
(i) the potential energy of the trolley at the top of the runway,
(ii) the kinetic energy of the trolley at the bottom of the runway,
(iii) the work done by the trolley against friction.

The trolley experiences a constant frictional force of 0.50 N throughout its motion, and takes 2.8 s to reach the bottom of the runway.

(b) Calculate
(i) the acceleration of the trolley,
(ii) the net force pulling the trolley down the runway,
(iii) the force of gravity pulling the trolley down the runway.

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Answer:

(a)
(i) Potential energy = mgh = 0.20 * 10 * 0.40 = 0.80 J

(ii) kinetic energy of trolley at bottom = ½ mv²
= ½ (0.20)(1.0)²
= 0.10 J

(iii) work done aganist friction = 0.80 J - 0.10 J = 0.70 J


(b)
(i) Acceleration of trolley = (1.0 m/s) / (2.8 s)
= 0.36 m/s²

(ii) Using F = ma,
net force = 0.20 * 0.36
= 0.072 N

(iii) Net force downwards = force of gravity down runway - frictional force
Force of gravity on trolley down runway = net force + frictional force
= 0.072 N + 0.50 N
= 0.57 N (2 s.f.)

O lvl Phy: Work Energy Power

A 4.0 kg mass of metal is dropped a distance 0.8m onto the horizontal surface in order to test its hardness.

(a) when the mass has fallen through 0.8m, how much gravitational potential energy has been transformed? [g = 10m/s²]

(b) How much kinetic energy does the mass possess before it hits the horizontal surface?

(c) What is the velocity of the body just before hitting the horizontal surface?

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Answer:

(a) Gravitational Potential Energy transformed
= mgh
= (4)(10)(0.8)
= 32 J


(b)
By the principle of conservation of energy,
Loss in PE = Gain in KE

Hence, KE = 32 J

(c) ½mv² = 32
v = √(32 * 2 / 4 )

v = 4 m/s

O lvl Phy: Work Energy Power

A ball of mass 4 kg slides down a circular slope of heigh 2 m as shown in the diagram. Assume no air resistance.


(a) State the type of energy the ball possesses prior to release from the top of the slope.

(b) Calculate the speed of the ball when it reaches point Q. Assume the slope is frictionless.

(c) If the track is not frictionless and the ball reaches point Q at 6 m/s. What is the work done against friction?

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Answer:

(a) Gravitational Potential Energy

(b) Using conservation of energy,
loss in GPE = gain in KE
mgh = ½ m v²
v² = 2gh
v² = 2 (10) (2) = 40
v = 6.32 m/s


(c) Using conservation of energy,
loss in GPE = gain in KE + work done against friction
mgh = ½ m v² + work done against friction
(4)(10)(2) = ½ (4) (6)² + work done against friction
work done against friction = 80 -72
work done against friction = 8 J

O lvl Phy: Work Energy Power

The figure below shows the path of a cart moving at various positions (A to E) along a smooth roller coaster. The 20.0 kg cart was projected at a height of 30.0 m above ground with an initial speed of 5.0 m/s.

(a) Calculate the speed at point B of the roller coaster ride.

(b) Calculate the maximum height (position C) the cart can reach. State the assumption made.

(c) Point D is the top of the circular loop. By indicating the force(s) acting on the cart at that point, explain why no work is done on the cart even though the cart is still moving at that point.

(d) Passengers have to be secured with safety belt. Throughout the ride, different forces will be exerted by the safety belt on the passengers to keep them on their seats. It is absolutely necessary at two points of the ride that the safety belt be working properly. State which are the 2 points and explain the choice of your points.

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Answer:

(a) Using the conservation of energy,
initial energy = final energy
½ m u² + mgh = ½ m v²
½ u² + gh = ½ v²
½ (5.0)² + (10)(30.0) = ½ v²
½ v² = 12.5 + 300
v² = 625
v = 25 m/s


(b) Using the conservation of energy,
initial energy = final energy
½ m v² = mgh
h = ½ v² / g
h = ½ (25)² / 10
h = 31.3 m

Assumptions:
1) No energy is lost due to air resistance
2) Cart reaches C with no speed.


(c)

The blue force is the force due to gravity.
The red force is the force due to the contact force from the path.

No work is done because the direction of motion is perpendicular to the direction of the force.


(d) Point C and Point E.

Point C because cart is moving down when inertia will move the body forward; the safety belt will pull the passenger down along with the cart as it moves down.

Point E because it is when the cart is stopping due to brakes. Inertia will cause the body to continue to move forward when the brakes are applied on the cart; the safety belt will exert a force to stop the passenger from being thrown forward out of the cart.

O lvl Phy: Work Energy Power

A crate of mass 80kg sits on the floor. the crate must be brought to the top of a loading dock by sliding it up a ramp 2.5m long, inclined at 30 degrees. the worker, giving no thought to the force of friction, calculates that he can get the crate up the ramp by giving it an initial speed of 5m/s at the bottom and letting it go. unfortunately, friction is not negligible, the crate slides up 1.6m on the ramp, stops, and slides back down.

Using conservation of energy, show that the initial speed of 5.0 m/s as calculated by the worker is correct if friction is negligible.

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Answer:

Initial kinetic energy = Final increase in potential energy

Because the ramp is angled at 30 degrees and is 2.5 m long, the vertical height increase is 2.5 sin 30 = 1.25m

1/2 m v² = mgh
1/2 m v² = m(10)(1.25)
v² = 25

Hence v = 5 m/s (shown)