Q: A car of mass 1300 kg accelerates from rest to 26.8 m/s in 10 s. If the engine provides a power of 230 hp ($$1\ \mathrm{hp} = 746\ \mathrm{W}$$), what is the work done by the engine?

$$W = \Delta KE = \frac{1}{2} m v^2$ = \frac{1}{2} \times 1300 \times (26.8)^2$ = 4.67 \times 10^5$\ \mathrm{J}$$

Q: A block of mass 8 kg is on the verge of sliding down a $$26^\circ$$ incline. The coefficient of static friction is 0.71. What is the magnitude of the static friction force?

$$f_{s,max} = \mu_s F_N = 0.71 \times 79.3 = 56.3\ \mathrm{N}$$

Q: A block of mass 14 kg is pulled to the left by a force $$F_1 = 80$$\ \mathrm{N}$$ at 25$$^\circ$$ above the horizontal. What is the vertical component of the pulling force?

$$F_{1y} = F_1$ \sin 25^\circ = 80 \times 0.4226 = 33.8\ \mathrm{N}$$

Q: A ball of mass 0.4 kg is attached to a string and moves in a circle of radius 0.37 m at a speed of 0.7 m/s. What is the tension in the string?

$$T = m \frac{v^2$}{r} = 0.4 \times \frac{(0.7)^2$}{0.37} = 0.53\ \mathrm{N}$$

Q: A block of mass 8 kg slides down a $$26^\circ$$ incline. The coefficient of kinetic friction is 0.61. What is the magnitude of the kinetic friction force?

$$f_k = \mu_k F_N = 0.61 \times 79.3 = 48.4\ \mathrm{N}$$

Question 1

A ball of mass 0.4 kg moves in a horizontal plane at a speed of 0.7 m/s. What is the magnitude of its centripetal acceleration if the radius of its circular path is 0.37 m?

Accepted Answer

$$a_c = \frac{v^2$}{r} = \frac{(0.7)^2$}{0.37} = 1.32\ \mathrm{m/s^2$}$$

Question 2

A sledder of mass 80 kg starts from rest at the top of a 32 m high hill and slides down to the top of a second hill 42 m away, reaching a speed of 8.4 m/s. Neglecting friction, what was the initial speed at the top of the first hill?

Accepted Answer

$$v_0 = 0$$\ \mathrm{m/s}$$

Question 3

A block of mass 14 kg is pulled to the left across a rough floor by a force $$F_1 = 80$$\ \mathrm{N}$$ at 25$$^\circ$$ above the horizontal. The kinetic friction force is F_k = 24$$\ \mathrm{N}$$. What is the net work done on the block after it moves 8.0 m?

Accepted Answer

$$W_{net} = 341 - 8F_k = 341 - 192 = 149\ \mathrm{J}$$

Question 4

A block of mass 8 kg slides down a $$26^\circ$$ incline. The coefficient of static friction is 0.71 and kinetic friction is 0.61. What is the magnitude of the normal force on the block?

Accepted Answer

$$F_N = 79.3\ \mathrm{N}$$

Question 5

A car of mass 1300 kg accelerates from rest to 26.8 m/s in 10 s. If the engine provides a power of 230 hp ($$1\ \mathrm{hp} = 746\ \mathrm{W}$$), what is the work done by the engine?

Accepted Answer

$$W = \Delta KE = \frac{1}{2} m v^2$ = \frac{1}{2} 	imes 1300 	imes (26.8)^2$ = 4.67 	imes 10^5$\ \mathrm{J}$$

Question 6

A spring with natural length 15 cm and spring constant $$k = 500\ \mathrm{N/m} is $$compressed to 10 cm. A block of mass 0.2 kg is attached and released. If a constant force $$F = 0.8\ \mathrm{N}$$ pulls the block to the right, what is the block's speed when the spring is at 25 cm?

Accepted Answer

$$v = \sqrt{\frac{1}{m} \left[ F \Delta x + \frac{1}{2} k (x_i^2$ - x_f^2$) ight]} = \sqrt{\frac{1}{0.2} \left[ 0.8 	imes 0.15 + \frac{1}{2} 	imes 500 	imes ((0.10)^2$ - (0.25)^2$) ight]}$$

Question 7

A 0.3 kg ball collides with a 0.7 kg ball and bounces back. If the initial velocity of the 0.3 kg ball is 15 m/s toward the 0.7 kg ball at rest, and after the collision the 0.3 kg ball moves at -7 m/s, what is the velocity of the 0.7 kg ball after the collision? (Assume elastic collision.)

Accepted Answer

$$v_{2f} = \frac{m_1$ (v_{1i} - v_{1f})}{m_2$} = \frac{0.3 (15 - (-7))}{0.7} = 9.43\ \mathrm{m/s}$$

Question 8

A block of mass 8 kg is on the verge of sliding down a $$26^\circ$$ incline. The coefficient of static friction is 0.71. What is the magnitude of the static friction force?

Accepted Answer

$$f_{s,max} = \mu_s F_N = 0.71 	imes 79.3 = 56.3\ \mathrm{N}$$

Question 9

A block of mass 8 kg slides down a $$26^\circ$$ incline under an applied force $$F = 60\ \mathrm{N}$$ parallel to the incline. The coefficient of kinetic friction is 0.61. What is the acceleration of the block?

Accepted Answer

$$a = \frac{38.7 - 48.4 + 60}{8} = 6.3\ \mathrm{m/s^2$}$$

Question 10

A block of mass 14 kg is pulled to the left by a force $$F_1 = 80$$\ \mathrm{N}$$ at 25$$^\circ$$ above the horizontal. What is the vertical component of the pulling force?

Accepted Answer

$$F_{1y} = F_1$ \sin 25^\circ = 80 	imes 0.4226 = 33.8\ \mathrm{N}$$

Question 11

A ball of mass 0.4 kg is attached to a string and moves in a circle of radius 0.37 m at a speed of 0.7 m/s. What is the tension in the string?

Accepted Answer

$$T = m \frac{v^2$}{r} = 0.4 	imes \frac{(0.7)^2$}{0.37} = 0.53\ \mathrm{N}$$

Question 12

A block of mass 8 kg slides down a $$26^\circ$$ incline. The coefficient of kinetic friction is 0.61. What is the magnitude of the kinetic friction force?

Accepted Answer

$$f_k = \mu_k F_N = 0.61 	imes 79.3 = 48.4\ \mathrm{N}$$

Question 13

A block of mass 14 kg is pulled to the left by a force $$F_1 = 80$$\ \mathrm{N}$$ at 25$$^\circ$$ above the horizontal. What is the horizontal component of the pulling force?

Accepted Answer

$$F_{1x} = F_1$ \cos 25^\circ = 80 	imes 0.9063 = 72.5\ \mathrm{N}$$

Question 14

A block of mass 8 kg slides down a $$26^\circ$$ incline. The coefficient of kinetic friction is 0.61. What is the net force acting on the block?

Accepted Answer

$$F_{net} = m g \sin 	heta - f_k = 8 	imes 9.8 	imes \sin 26^\circ - 48.4 = 34.4 - 48.4 = -14.0\ \mathrm{N}$$

Question 15

A block of mass 14 kg is pulled to the left by a force $$F_1 = 80$$\ \mathrm{N}$$ at 25$$^\circ$$ above the horizontal. What is the work done by the pulling force after the block moves 8.0 m?

Accepted Answer

$$W = F_1$ \cos 25^\circ 	imes 8.0 = 80 	imes 0.9063 	imes 8.0 = 581\ \mathrm{J}$$

Question 16

A block of mass 14 kg is pulled to the left by a force $$F_1 = 80$$\ \mathrm{N}$$ at 25$$^\circ$$ above the horizontal. The kinetic friction force is F_k = 24$$\ \mathrm{N}$$. What is the total work done on the block after it moves 8.0 m?

Accepted Answer

$$W_{total} = F_1$ \cos 25^\circ 	imes 8.0 - F_k 	imes 8.0 = 581 - 192 = 389\ \mathrm{J}$$

Physics Exam Study Guide: Forces, Work, Energy, and Motion

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