Question

1. Calculate the work done by a 2.0-N force (directed at a 80° angle to the vertical) to move a 500 gram box a horizontal distance of 500 cm across a rough floor at a constant speed of 0.8 m/s. Be cautious with the units. Show your solution.

2. Mrs. Salcedo's car of mass 2.4 x 10³ kg travelling on the highway when the teacher receives an emergency call. She increased the speed of the car to 33 m/s. The increase in speed results in 3.1 x10³ J of work done on the car. Determine the initial speed of the car in km/h.

3. A 80.0-kg person jumps onto the floor from a height of 5.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints.

Answer 1

(1) By the definition of the work done, we get:

$$W=Fdcos\theta,$$ $$W=2.0\ N\times500\ cm\times\dfrac{1\ m}{100\ cm}\times cos(90^{\circ}-80^{\circ})=9.85\ J.$$

(2) We can find the initial speed of the car from the work-kinetic energy theorem:

$$W=\Delta KE=KE_f-KE_i,$$ $$W=\dfrac{1}{2}mv_f^2-\dfrac{1}{2}mv_i^2,$$ $$v_i=\sqrt{\dfrac{2(\dfrac{1}{2}mv_f^2-W)}{m}},$$ $$v_i=\sqrt{\dfrac{2\times(\dfrac{1}{2}\times2.4\times10^3\ kg\times(33\ \dfrac{m}{s})^2-3.1\times10^3\ J)}{2.4\times10^3\ kg}},$$ $$v_i=32.96\ \dfrac{m}{s}\times\dfrac{1\ km}{1000\ m}\times\dfrac{3600\ s}{1\ h}=118.6\ \dfrac{km}{h}.$$

(3) By the definition of the work done, we get:

$$W=Fd.$$

From the other hand, using the work-kinetic energy theorem, we have:

$$W=\Delta KE=mgh.$$

Equating these two expressions, we get:

$$Fd=mgh,$$ $$F=\dfrac{mgh}{d}=\dfrac{80\ kg\times9.8\ \dfrac{m}{s^2}\times5\ m}{5\times10^{-3}\ m}=7.84\times10^5\ N.$$