21. A forceof 5.0 N moves a 6.0 kg object along a rough floor at a constant spee

Question

21. A forceof 5.0 N moves a 6.0 kg object along a rough floor at a constant speed of 2.5 m/s.
(a) How much work is done in 25 s?
(b) What power is being used?
(C) What force of friction is acting on the object?

22. A 3.0kg metal ball, at rest, is hit by a 1.0 kg metal ball moving at 4.0 m/s. The 3.0 kg ball moves forward at 2.0 m/s and the 1.0 kg ball bounces back at 2.0 m/s.
a. what is the total kE before collision
b. what is the total KE after collision
c. how much energy is transferred fromt he small ball to the large ball?

Formulas we use: PE= mgh KE = 1/2mv^2 w=fxd p= w/t Mechanical Eff. WO/Wi x100
gravity= 10

Explanation / Answer

a) Work = force x distance

    Work = 5 x ( 2.5x25)=312.5 J

b)Power = work/time= 312.5 /25= 12.5 W

c) Force of friction = 5N ( as the body is moving with constant velocioty)

e)KE before colloision= KE of 1 kg + KE of 3 kg

                                   = 0.5(1)42 + 0= 8J

f)Total kE afer collosion= KE of 1 kg + KE of 3 Kg

                                     0.5(1)22 +0.5(3)22

                                        =2 + 6 = 8 J

g) KE transferred = 8J - 2J = 6J

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