A box of mass 18.3 kg with an initial velocity of 1.88 m/s slides down a plane,

Question

A box of mass 18.3 kg with an initial velocity of 1.88 m/s slides down a plane, inclined at 28 with respect to the horizontal. The coefficient of kinetic friction is 0.77. The box stops after sliding a distance x. How far does the box slide? The acceleration due to gravity is 9.8 m/s 2 . The positive x-direction is down the plane. Answer in units of m.

Part B: What is the the work done by friction? Answer in units of J.

Part C: What is the work done by the normal force? Answer in units of J.

Part D: What is the magnitude of the work done by gravity? Answer in units of J.

Part E: What is the magnitude of the instantaneous power generated by friction half way between the initial and final positions? Answer in units of W.

Part F: What is the magnitude of the average power generated by friction from start to stop? Answer in units of W.

Explanation / Answer

Resolving forces, acting upon the block, into components we get

N (normal reaction) = mg cos (theta)

Fs (frictional force) = mg sin (theta) + ma

(a). F = mg sin (theta) - Fs

ma = mg sin (theta) - (Mu) N

ma = mg sin(theta) - Mu mg cos(theta)

a = g sin (theta) - Mu g cos(theta)

a = 9.8 [ sin(28) - 0.77 cos(28) ]

a = -2.06094 meter per second square (negative sign indicates retardation)

As the block stops after covering a certain distance x, final velocity v = 0

v = u +at

0 = 1.88 - 2.06094 t

t = 1.88 / 2.06094 = 0.912 seconds

Thus, x = ut + (a*t*t)/2

x = (1.88*0.912) - ((2.06094*0.912*0.912)/2)

x = 1.71456 - 0.8571

x = 0.85746 meters.

(b) Work done by friction = frictional force * displacement

= (Mu)N * x

= (Mu) mg cos (theta) x

= 0.77*18.3*9.8*0.85746*cos(28)

= 104.49 Joules

(c) Work done by normal force = normal force * displacement

= N*x

= mgx cos(theta)

= 18.3*9.8*0.85746 cos(28)

= 135.704 Joules

(d) Work done by gravity = mg sin(theta) x

= 18.3*9.8*0.85746* sin(28)

= 72.155 Joules

(e) for half way x/2 = 0.42873

Now, s = ut + (a*t*t)/2

0.42873 = 1.88t - (2.06094/2)*t*t

thus, t = 0.267 sec

Therefore, Power = work / time

P = 104.49 / 0.267

P = 391.35 Watt

(f) Power = 104.49 / 0.912

P = 114.57 Watt

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