Question

During a storm, a crate of crepe is sliding across a slick, oily parking lot through a displacement d=(-3.0m)i while a steady wind pushes against the crate with a force F=(2.0N)i+(-6.0N)j. (a) How much work does this force do on the crate during the displacement? (b) If the crate has a kinetic energy of 10J at the beginning of displacement d, what is its kinetic energy at the end of d?

Answer 1

Answer on Question #62329, Physics / Mechanics | Relativity

During a storm, a crate of crepe is sliding across a slick, oily parking lot through a displacement $d = (-3.0m)i$ while a steady wind pushes against the crate with a force $F = (2.0N)i + (-6.0N)j$ .

(a) How much work does this force do on the crate during the displacement?

(b) If the crate has a kinetic energy of 10J at the beginning of displacement $d$ , what is its kinetic energy at the end of $d$ ?

Solution:

(a)

Because we can treat the crate as a particle and because the wind force is constant ("steady") in both magnitude and direction during the displacement to calculate the work we can use equation

$$W = \vec {F} \cdot \vec {d}$$ $$\begin{array}{l} W = \left[ (2. 0 \mathrm {N}) \vec {\mathrm {i}} + (- 6. 0 \mathrm {N}) \vec {\mathrm {j}} \right] \cdot \left[ (- 3. 0 \mathrm {m}) \vec {\mathrm {i}} \right] = \\ = (2. 0 \mathrm {N}) (- 3. 0 \mathrm {m}) \vec {\mathrm {i i}} + (- 6. 0 N) (- 3. 0 m) \vec {\mathrm {i j}} = \\ = (- 6. 0 \mathrm {J}) (1) + 0 = - 6. 0 \mathrm {J} \\ \end{array}$$

Thus, the force does a negative 6.0 J of work on the crate, transferring 6.0 J of energy from the kinetic energy of the crate.

(b) Because the force does negative work on the crate, it reduces the crate's kinetic energy.

Using the work-kinetic energy theorem we have

$$K _ {f} = K _ {i} + W = 1 0 \mathrm {J} + (- 6. 0 \mathrm {J}) = 4. 0 \mathrm {J}.$$

Less kinetic energy means that the crate has been slowed.

Answer: (a) -6.0 J;

(b) 4.0 J.

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