Question

unit 6 - energy practice test

1. a 5 kg box is moved from the floor to a shelf that is 2.5 m high.

a. how much work was done putting the box on the shelf?
b. find the change in gravitational energy.

1. a driver in a honda civic notices traffic congestion up ahead. he slows down from 30.0 m/s to 15 m/s. assume the mass of the driver and car is 1450 kg.

a. calculate the change in kinetic energy.
b. how much energy was dissipated in the process?

1. a spring k = 470 n/m in a toy dart gun is compressed 0.025 m when loaded. assuming the dart has a mass of 0.015 kg, how much elastic energy is stored in the spring loaded dart gun.
2. nellie newton lifts a 50 kg box 3 m in 5 seconds. how much power does she exert in doing this?
3. net force as a function of displacement of a moving object is presented by the graph below. how much work is done when the object moves from 0 m to 30 m?
4. the diagram to the right depicts forces acting on a 2 kg block to slide a distance of 4 m across a classroom floor. the work done is 25 j, and the force of friction is 10 n. what is the applied force?

a. friction?
b. gravity?
c. the normal force?
d. the applied force?
e. the net force?

Explanation:

Question 6

Step1: Define work done

Work done against gravity: $W = mgh$
$m=5\ \text{kg}, g=9.8\ \text{m/s}^2, h=2.5\ \text{m}$

Step2: Calculate work

$W = 5 \times 9.8 \times 2.5 = 122.5\ \text{J}$

Step3: Relate to gravitational PE

Change in gravitational potential energy equals work done: $\Delta PE = W = 122.5\ \text{J}$

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Question 7

Step1: Kinetic energy formula

$KE = \frac{1}{2}mv^2$
$m=1450\ \text{kg}, v_1=30.0\ \text{m/s}, v_2=15\ \text{m/s}$

Step2: Calculate initial KE

$KE_1 = \frac{1}{2} \times 1450 \times 30.0^2 = 6.525 \times 10^5\ \text{J}$

Step3: Calculate final KE

$KE_2 = \frac{1}{2} \times 1450 \times 15^2 = 1.63125 \times 10^5\ \text{J}$

Step4: Find KE change

$\Delta KE = KE_2 - KE_1 = 1.63125 \times 10^5 - 6.525 \times 10^5 = -4.5 \times 10^5\ \text{J}$

Step5: Dissipated energy

Magnitude of $\Delta KE$ is dissipated: $4.5 \times 10^5\ \text{J}$

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Question 8

Step1: Elastic PE formula

$PE_{spring} = \frac{1}{2}kx^2$
$k=670\ \text{N/m}, x=0.025\ \text{m}$

Step2: Calculate stored energy

$PE_{spring} = \frac{1}{2} \times 670 \times (0.025)^2 = 0.5 \times 670 \times 0.000625 = 0.107\ \text{J}$

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Question 9

Answer:

a. 122.5 J
b. 122.5 J