6. A 10 kg block, initially at rest, is pulled 4.00 meters along an incline angl

Question

6. A 10 kg block, initially at rest, is pulled 4.00 meters along an incline angled 30 degree from horizontal by a force of 75.0 newtons parallel to the incline. The coefficient of kinetic friction between the block and incline is 0.20. For the 4.00 meter displacement, find: a) the force of friction opposing the motion, b) the non-conservative work done by the friction force, c) the non-conservative work done by the applied 75.0 N force, d) the change in gravitational potential energy of the block, e) the change in kinetic energy of the block, f) the speed of the block when it has been pulled 4.00 meters, and, g) the instantaneous power supplied by the 75.0 N force at that time.

Explanation / Answer

a) F(friction) = muek*m*g*cos(30)

= 0.2*10*9.8*cos(30)

= 3.02 N

b) W(fristion) = Ff*d*cos(180)

= -3.02*4

= -12.08 J

c) W(apllied) = F*d*cos(0)

= 75*4

= 300 J

d) delta U = m*g*d*sin(30)

= 10*9.8*4*sin(30)

= 196 J

e) Wnet = K2 - K1

W(friction) + W(applied force) + W(gravity) = K2 - K1

K2 - K1 = -12.08 + 300 - 19.6

= 91.92 J

f) K2 - K1 = 0.5*m*(v2^2-v1^2)

here v1 = 0

v2 = sqrt(2*(k2-k1)/m)

= sqrt(2*91.92/10)

= 4.28 m/s

g) P = F*v

= 75*4.28

= 32.16 Watts

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