Q: A 0.60 kg particle has a speed of 2.00 m/s at point A and a kinetic energy of 7.50 J at point B. What is the speed at point B?

$$v_B = \sqrt{\frac{2 \times 7.50}{0.60}} = 5.0\ \mathrm{m/s}$$

Q: A 0.300 kg ball has a speed of 15.0 m/s. What is its kinetic energy?

$$KE = \frac{1}{2} m v^2$ = \frac{1}{2} \times 0.300 \times (15.0)^2$ = 33.75\ \mathrm{J}$$

Q: If the speed of the 0.300 kg ball in the previous question were doubled, what would be its new kinetic energy?

$$KE_{new} = \frac{1}{2} \times 0.300 \times (30.0)^2$ = 135\ \mathrm{J}$$

Q: A 3.00 kg mass has an initial velocity $$v_i = 6.00\ \mathrm{m/s}$$ and a final velocity $$v_f = 4.00\ \mathrm{m/s}. $$What is the total work done on the object?

$$W = \frac{1}{2} m (v_f^2$ - v_i^2$) = \frac{1}{2} \times 3.00 \times (16 - 36) = -30\ \mathrm{J}$$

Q: A bullet with a mass of 5.00 g and a speed of 600 m/s penetrates a tree to a depth of 4.00 cm. Using energy considerations, what is the average frictional force that stops the bullet?

$$F = \frac{\frac{1}{2} m v^2$}{d} = \frac{0.5 \times 0.005 \times 600^2$}{0.04} = 22,500\ \mathrm{N}$$

Q: A block of mass 0.80 kg is given an initial velocity $$v_A = 1.2\ \mathrm{m/s} to $$the right and collides with a spring of force constant $$k = 50\ \mathrm{N/m}. $$Assuming a frictionless surface, what is the maximum compression of the spring?

$$x_{max} = \sqrt{\frac{m v_A^2$}{k}} = \sqrt{\frac{0.80 \times (1.2)^2$}{50}} = 0.152\ \mathrm{m}$$

Q: A particle of mass 0.500 kg is shot from point P with a horizontal velocity of 50.0 m/s and reaches a maximum height of 20.0 m above P. What is the vertical component of its initial velocity?

$$v_{y0} = \sqrt{2 g h} = \sqrt{2 \times 9.8 \times 20.0} = 19.8\ \mathrm{m/s}$$

Q: For the projectile in the previous question, what is the total time of flight from P to B?

$$t = \frac{2 v_{y0}}{g} = \frac{2 \times 19.8}{9.8} = 4.04\ \mathrm{s}$$

Q: A picture tube in a television set is 36.0 cm long. The electron inside accelerates from rest in the tube from rest to 10.0% of the speed of light. What is the kinetic energy of the electron at the end of the tube? (Electron mass $$m_e = 9.11 \times 10$$^{-31}$$\ \mathrm{kg}$$, $$c = 3.00 \times 10^8$\ \mathrm{m/s})$$

$$KE = \frac{1}{2} m_e v^2$ = \frac{1}{2} \times 9.11 \times 10$$^{-31}$$ \times (3.00 \times 10^7$)^2 = 4.10$$ \times $$10^{-16}$$\ \mathrm{J}$

Q: A 3.00 kg mass has an initial velocity of 6.00 m/s. What is its kinetic energy at this time?

$$KE = \frac{1}{2} m v^2$ = \frac{1}{2} \times 3.00 \times (6.00)^2$ = 54.0\ \mathrm{J}$$

Question 1

A 0.60 kg particle has a speed of 2.00 m/s at point A and a kinetic energy of 7.50 J at point B. What is the speed at point B?

Accepted Answer

$$v_B = \sqrt{\frac{2 	imes 7.50}{0.60}} = 5.0\ \mathrm{m/s}$$

Question 2

A 0.300 kg ball has a speed of 15.0 m/s. What is its kinetic energy?

Accepted Answer

$$KE = \frac{1}{2} m v^2$ = \frac{1}{2} 	imes 0.300 	imes (15.0)^2$ = 33.75\ \mathrm{J}$$

Question 3

If the speed of the 0.300 kg ball in the previous question were doubled, what would be its new kinetic energy?

Accepted Answer

$$KE_{new} = \frac{1}{2} 	imes 0.300 	imes (30.0)^2$ = 135\ \mathrm{J}$$

Question 4

A 3.00 kg mass has an initial velocity $$v_i = 6.00\ \mathrm{m/s}$$ and a final velocity $$v_f = 4.00\ \mathrm{m/s}. $$What is the total work done on the object?

Accepted Answer

$$W = \frac{1}{2} m (v_f^2$ - v_i^2$) = \frac{1}{2} 	imes 3.00 	imes (16 - 36) = -30\ \mathrm{J}$$

Question 5

A bullet with a mass of 5.00 g and a speed of 600 m/s penetrates a tree to a depth of 4.00 cm. Using energy considerations, what is the average frictional force that stops the bullet?

Accepted Answer

$$F = \frac{\frac{1}{2} m v^2$}{d} = \frac{0.5 	imes 0.005 	imes 600^2$}{0.04} = 22,500\ \mathrm{N}$$

Question 6

A block of mass 0.80 kg is given an initial velocity $$v_A = 1.2\ \mathrm{m/s} to $$the right and collides with a spring of force constant $$k = 50\ \mathrm{N/m}. $$Assuming a frictionless surface, what is the maximum compression of the spring?

Accepted Answer

$$x_{max} = \sqrt{\frac{m v_A^2$}{k}} = \sqrt{\frac{0.80 	imes (1.2)^2$}{50}} = 0.152\ \mathrm{m}$$

Question 7

A particle of mass 0.500 kg is shot from point P with a horizontal velocity of 50.0 m/s and reaches a maximum height of 20.0 m above P. What is the vertical component of its initial velocity?

Accepted Answer

$$v_{y0} = \sqrt{2 g h} = \sqrt{2 	imes 9.8 	imes 20.0} = 19.8\ \mathrm{m/s}$$

Question 8

For the projectile in the previous question, what is the total time of flight from P to B?

Accepted Answer

$$t = \frac{2 v_{y0}}{g} = \frac{2 	imes 19.8}{9.8} = 4.04\ \mathrm{s}$$

Question 9

A picture tube in a television set is 36.0 cm long. The electron inside accelerates from rest in the tube from rest to 10.0% of the speed of light. What is the kinetic energy of the electron at the end of the tube? (Electron mass $$m_e = 9.11 	imes 10$$^{-31}$$\ \mathrm{kg}$$, $$c = 3.00 	imes 10^8$\ \mathrm{m/s})$$

Accepted Answer

$$KE = \frac{1}{2} m_e v^2$ = \frac{1}{2} 	imes 9.11 	imes 10$$^{-31}$$ 	imes (3.00 	imes 10^7$)^2 = 4.10$$ 	imes $$10^{-16}$$\ \mathrm{J}$

Question 10

A 3.00 kg mass has an initial velocity of 6.00 m/s. What is its kinetic energy at this time?

Accepted Answer

$$KE = \frac{1}{2} m v^2$ = \frac{1}{2} 	imes 3.00 	imes (6.00)^2$ = 54.0\ \mathrm{J}$$

Question 11

A bullet of mass 5.00 g is fired at 600 m/s into a tree and stops after penetrating 4.00 cm. How much time elapses between the moment the bullet enters the tree and the moment it stops?

Accepted Answer

$$t = \frac{2d}{v} = \frac{2 	imes 0.04}{600} = 1.33 	imes 10$$^{-4}$$\ \mathrm{s}$$

Question 12

A block of mass 0.80 kg slides on a frictionless surface and compresses a spring with force constant $$k = 50\ \mathrm{N/m}. If $$the block's initial kinetic energy is 0.576 J, what is the maximum compression of the spring?

Accepted Answer

$$x_{max} = \sqrt{\frac{2 KE}{k}} = \sqrt{\frac{2 	imes 0.576}{50}} = 0.152\ \mathrm{m}$$

Question 13

A 0.500 kg particle is projected horizontally at 50.0 m/s from a height of 40.0 m. Neglecting air resistance, what is the horizontal distance traveled before it hits the ground?

Accepted Answer

$$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 	imes 40.0}{9.8}} = 2.86\ \mathrm{s}$$, $$d = v_x t = 50.0 	imes 2.86 = 143\ \mathrm{m}$$

Question 14

A 0.60 kg particle moves from point A to B. If the work done on the particle is 5.0 J and its initial kinetic energy at A is 1.2 J, what is its kinetic energy at B?

Accepted Answer

$$KE_B = KE_A + W = 1.2 + 5.0 = 6.2\ \mathrm{J}$$

Question 15

A 3.00 kg mass changes its velocity from 6.00 m/s to 4.00 m/s. What is the change in its kinetic energy?

Accepted Answer

$$\Delta KE = \frac{1}{2} m (v_f^2$ - v_i^2$) = \frac{1}{2} 	imes 3.00 	imes (16 - 36) = -30\ \mathrm{J}$$

Question 16

A 0.500 kg particle is projected at an angle such that its horizontal velocity is 50.0 m/s and it reaches a maximum height of 20.0 m. What is the total mechanical energy at the highest point?

Accepted Answer

$$E = KE_x + PE = \frac{1}{2} m v_x^2$ + mgh = 0.25 	imes 2500 + 0.5 	imes 9.8 	imes 20 = 625 + 98 = 723\ \mathrm{J}$$

Work, Energy, and Power: Homework Problems and Applications

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