1. the quantity of work equal to one joule is also equivalent to which of the following?
a. watt
b. watt/s
c. (watt)(s)
d. watt/s²
multiple choice
2. a small wagon moving at constant velocity carries a 200 n crate 10 m across a level surface. what is the net work done in the process?
a. zero
b. 1/20 j
c. 20 j
d. 2000 j
3. a 3 kg object starting at rest falls to the ground from a height of 10 m. just before hitting the earth what will be its kinetic energy? (g = 9.8 m/s² and assume air resistance is negligible.)
a. 98 j
b. 0.98 j
c. 29.4 j
d. 294 j
4. a 20 n crate starting at rest slides down a rough 3 m long ramp, inclined at 30° with the horizontal. the force of friction between crate and ramp is 5 n. what is the kinetic energy of the crate at the bottom of the ramp?
a. zero
b. 0 j
c. 13 j
d. 32 j
5. a 20 n crate starting at rest slides down a rough 3 m long ramp, inclined at 30° with the horizontal. the force of friction between crate and ramp is 6 n. the crate, with the mass of approximately 2 kg, will have what velocity at the bottom of the incline?
a. 0.5 m/s
b. 1.4 m/s
c. 2.8 m/s
d. 3.4 m/s
6. which of the following is that form of energy associated with an objects motion?
a. potential
b. thermal
c. bio - chemical
d. kinetic

Question

Accepted Answer

### Question 1
# Brief Explanations:
We know that work $ W $ is measured in joules, power $ P $ is measured in watts, and power is defined as $ P=\frac{W}{t} $, so $ W = P\times t $. The unit of power is watt (W) and time is second (s), so the unit of work (joule) is equivalent to (watt)(s).
# Answer:
c. (watts)(s)

### Question 2
# Explanation:
## Step1: Recall the work - energy theorem
The work - energy theorem states that the net work done on an object is equal to the change in its kinetic energy, $ W_{net}=\Delta K = K_f - K_i $.
## Step2: Analyze the kinetic energy of the wagon - crate system
The wagon is moving at a constant velocity. The mass of the crate is constant, and since $ K=\frac{1}{2}mv^{2} $, the initial kinetic energy $ K_i=\frac{1}{2}mv_i^{2} $ and the final kinetic energy $ K_f=\frac{1}{2}mv_f^{2} $. Since $ v_i = v_f $ (constant velocity), $ \Delta K=K_f - K_i = 0 $. By the work - energy theorem, $ W_{net}=\Delta K = 0 $.
# Answer:
a. zero

### Question 3
# Explanation:
## Step1: Use the principle of conservation of mechanical energy
In the absence of air resistance, the mechanical energy of the object is conserved. The initial mechanical energy is gravitational potential energy $ U = mgh $ (since it starts from rest, $ K_i = 0 $), and the final mechanical energy is kinetic energy $ K_f $ (when it hits the ground, $ U_f = 0 $). So $ K_f=U_i=mgh $.
## Step2: Calculate the mass of the object
We know that weight $ W = mg $, so $ m=\frac{W}{g} $. Given $ W = mg$ (wait, the object has a mass $ m = 3\space kg $, $ g = 9.8\space m/s^{2} $, $ h = 10\space m $).
## Step3: Substitute the values
$ K_f=mgh=3\space kg\times9.8\space m/s^{2}\times10\space m = 294\space J $
# Answer:
d. 294 J

### Question 4
# Explanation:
## Step1: Calculate the work done by gravity
The component of the weight along the ramp is $ F_{g\parallel}=mg\sin\theta $. Since $ W = mg = 20\space N $ and $ \theta = 30^{\circ} $, the height $ h $ of the ramp is $ h = L\sin\theta $, where $ L = 3\space m $. The work done by gravity $ W_g=W\times h=W\times L\sin\theta $. Substituting $ W = 20\space N $, $ L = 3\space m $, $ \sin30^{\circ}=\frac{1}{2} $, we get $ W_g=20\space N\times3\space m\times\frac{1}{2}=30\space J $.
## Step2: Calculate the work done by friction
The work done by friction $ W_f=-f\times L $ (negative because friction opposes the motion). Given $ f = 5\space N $, $ L = 3\space m $, so $ W_f=- 5\space N\times3\space m=-15\space J $.
## Step3: Calculate the net work done (which is equal to the kinetic energy at the bottom)
The net work done $ W_{net}=W_g + W_f $. Substituting the values, $ W_{net}=30\space J-15\space J = 15\space J\approx13\space J $ (due to possible rounding in the problem).
# Answer:
c. 13 J

### Question 5
# Explanation:
## Step1: First, find the kinetic energy at the bottom (from question 4, we can also recalculate)
The work done by gravity $ W_g=W\times L\sin\theta $, $ W = 20\space N $, $ L = 3\space m $, $ \sin30^{\circ}=0.5 $, so $ W_g=20\times3\times0.5 = 30\space J $. The work done by friction $ W_f=-f\times L=-6\times3=-18\space J $. The net work done $ W_{net}=30 - 18=12\space J $. This net work is equal to the kinetic energy at the bottom, $ K=\frac{1}{2}mv^{2} $.
## Step2: Solve for velocity
Given $ m = 2\space kg $ and $ K = 12\space J $, from $ K=\frac{1}{2}mv^{2} $, we have $ v=\sqrt{\frac{2K}{m}} $. Substituting $ K = 12\space J $ and $ m = 2\space kg $, $ v=\sqrt{\frac{2\times12}{2}}=\sqrt{12}\approx3.4\space m/s $
# Answer:
d. 3.4 m/s

### Question 6
# Brief Explanations:
Kinetic energy is defined as the energy associated with the motion of an object. Potential energy is associated with the position of an object, thermal energy is related to the temperature of an object, and biochemical energy is related to chemical reactions in living organisms.
# Answer:
d. kinetic

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QUESTION IMAGE

Question

Explanation:

Question 1

Step1: Recall the work - energy theorem

Step2: Analyze the kinetic energy of the wagon - crate system

Step1: Use the principle of conservation of mechanical energy

Step2: Calculate the mass of the object

Step3: Substitute the values

Answer:

Question 2