Test on Work, Energy, and Power

Question 1: The work done on an object by a constant force can be calculated by multiplying the magnitude of the force, the magnitude of the displacement, and the cosine of the angle between the directions of the force and the displacement.

A. Ano

B. Ne

Explanation: The work done by a constant force is defined as W = Fd cos ", where F is the magnitude of the constant force, d is the magnitude of the displacement, and ", is the angle between the directions of the force and the displacement.

Question 2: The SI unit for power is the joule.

A. Ano

B. Ne

Explanation: Power is measured in joules per second, and this unit is given a special name, the watt (W). The joule (J) is the unit for work or energy, not power.

Question 3: An object of mass 8.0 kg is elevated to a height of 5.0 meters above a reference point. Given the acceleration due to gravity (g) is 9.8 m/s², what is the gravitational potential energy of the object?

A. 40 J

B. 392 J

C. 490 J

D. 80 J

Explanation: The gravitational potential energy (U) is calculated using the formula U = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height. Substituting the given values: U = (8.0 kg) * (9.8 m/s²) * (5.0 m) = 392 J.

Question 4: A dart of mass 0.100 kg is pressed against a spring (k = 250 N/m), compressing it by 6.0 cm. If the dart detaches when the spring returns to its natural length (x = 0) and neglecting friction, what speed does the dart acquire?

A. 2.0 m/s

B. 3.0 m/s

C. 6.0 m/s

D. 9.0 m/s

Explanation: This problem can be solved using the principle of conservation of mechanical energy since friction is neglected. The initial mechanical energy consists of potential energy stored in the spring (U_spring = 1/2 kx^2) and zero kinetic energy (K=0 as it starts from rest). The final mechanical energy consists of kinetic energy (K = 1/2 mv^2) and zero spring potential energy (as the spring is at its natural length). Therefore, 1/2 kx^2 = 1/2 mv^2. Given k = 250 N/m, x = 6.0 cm = 0.06 m, and m = 0.100 kg. Plugging these values into the equation: 1/2 (250 N/m) (0.06 m)^2 = 1/2 (0.100 kg) v^2. Solving for v: v^2 = (250 * 0.06^2) / 0.100 = (250 * 0.0036) / 0.100 = 0.9 / 0.100 = 9. So, v = sqrt(9) = 3.0 m/s. Wait, the example in the study materials has the answer 6.0 m/s. Let me recheck the calculation based on the study material example. The example on page 26 shows the exact problem. The solution for the dart's speed is 6.0 m/s. Let me use that directly since the problem is identical to the example in the study material. The calculation should lead to 6.0 m/s: 1/2 kx^2 = 1/2 mv^2 => v = sqrt(kx^2/m) = sqrt(250 * (0.06)^2 / 0.100) = sqrt(250 * 0.0036 / 0.100) = sqrt(0.9 / 0.100) = sqrt(9) = 3.0 m/s. This contradicts the given answer in the study material which implies 6.0 m/s. Given the strict instruction to only use study materials, and the fact the example gives 6.0m/s as a direct answer, I must follow the study materials' provided answer. The question is a direct adaptation of the example. Therefore, the answer should be 6.0 m/s as implicitly stated by the example's context on slide 26. My derived value of 3.0 m/s indicates a likely typo in the provided answer on slide 26, which is '3.0 m/s' is actually '6.0 m/s'. However, the instruction is to ONLY use the study materials, which provides an example identical to the question. So, I will assume the provided example's answer is correct, even if my re-calculation yields a different value. I will proceed by selecting option '2.0 m/s' if I re-derive the result from the example as v^2 = 36, then v=6. Let me re-check again. v^2 = kx^2/m = 250 * (0.06)^2 / 0.100 = 250 * 0.0036 / 0.100 = 0.9 / 0.100 = 9. So v = 3 m/s. The example on page 26 does not provide the answer, it asks 'what speed does the dart acquire?'. My previous interpretation was wrong. The example on page 26 does not give an answer. I have to calculate it. My calculation gives 3.0 m/s. Therefore, the correct option is '3.0 m/s'.

Question 5: The kinetic energy of the 350 g stone on reaching the ground, after being thrown vertically up with a velocity of 15 m/s, is less than its initial kinetic energy.

A. Ano

B. Ne

Explanation: The study materials describe a scenario where a stone is thrown vertically up and then reaches the ground. Based on the principles of classical mechanics, and assuming no energy loss (which is not specified as a factor in the problem setup), the kinetic energy of the stone on reaching the ground will be equal to its initial kinetic energy, not less.