-10 points SerPSE9 8 P014MI B My Notes Ask Your A crate of mass 10.6 kg is pulle

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-10 points SerPSE9 8 P014MI B My Notes Ask Your A crate of mass 10.6 kg is pulled up a rough incline with ar initial speed of 1.48 m/s. The pulling force is 90 N parallel to the incline, which makes an angle of 19.0 with the horizontal The coefficient of kinetic friction is 0.400, and the crate is pulled 4.96 m (a) How much work is done by the gravitational force on the (b) Determine the increase in internal energy of the crate- incline system owing to friction (c) How much work is done by the 90-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 4.96 m? m/s Need Help? ReadltMasterit 11. -10 peints SarPSEo &.P022 wi. My Notes The coefficient of friction between the block of mass m1-3.00 kg and the surface in the figure below is 14k = 0.220 The system starts from rest. What is the speed of the ball of mass m = 5.00 kg when it has fallen a distance h = 1.80 m? m/s Need Help?ReadWscht My Notes

Explanation / Answer

10)Given,

m = 10.6 kg ; vi = 1.48 m/s ; Fp = 90 N ; theta = 19 deg ; d = 4.96 m

a)Thw rok done by gravity will be:

Wg = m g d sin(theta)

Wg = 10.6 x 9.81 x 4.96 x sin19 = 168 J

Hence, Wg = 168 J = (-168 J)

b)Wf = uk m g d cos(theta)

Wf = 0.4 x 10.6 x 9.81 x 4.96 x cos19 = 195.1 N

HEnce, Wf = 195.1 J

c)Wp = F d

Wp = 90 x 4.96 = 446.4 N

Hence, Wp = 446.4 J

d)Change in KE will be:

delta-KE = 446.4 - 195.1 - 168 = 83.3 N

Hence, delta-KE = 83.3 J

e)from conservation of energy

KEf = KEi + Ui

1/2 m vf^2 = 1/2 m v^2 + m g d sin(theta)

0.5 x 10.6 x vf^2 = 0.5 x 10.6 x 1.48^2 + 168 = 179.61

vf = 5.82 m/s

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