Question

A mass m = 17 kg is pulled along a horizontal floor with NO friction for a distance d =5.7 m. Then the mass is pulled up an incline that makes an angle θ = 34° with the horizontal and has a coefficient of kinetic friction μk = 0.38. The entire time the massless rope used to pull the block is pulled parallel to the incline at an angle of θ = 34° (thus on the incline it is parallel to the surface) and has a tension T =83 N.

What is the work done by tension before the block goes up the incline? (On the horizontal surface.)

What is the speed of the block right before it begins to travel up the incline?

What is the work done by friction after the block has traveled a distance x = 2.6 m up the incline? (Where x is measured along the incline.)

What is the work done by gravity after the block has traveled a distance x = 2.6 m up the incline? (Where x is measured along the incline.)

How far up the incline does the block travel before coming to rest? (Measured along the incline.)

Answer 1

Part A.

Work-done is given by:

W1 = F.d = F*d*cos

Force done by tension before block goes up the incline will be:

= angle between force applied and displacement = 34 deg

T = Tension applied = 83 N

d = displacement on horizontal surface = 5.7 m

So,

W1 = 83*5.7*cos 34 deg

W1 = 392.2 J

Part B.

Using Work-energy theorem:

W1 = dKE

W1 = KEf - KEi

KEi = 0, since block starts from rest

KEf = (1/2)*m*V^2

V = speed of block right before it begins to travel up the incline = ?

W1 = (1/2)*m*V^2 - 0

V = sqrt (2*W1/m)

V = sqrt (2*392.2/17) = 6.79 m/sec

Part C.

Work-done by friction force up the incline will be:

Wf = Ff*d1*cos

Ff = Force of friction along incline = *N = *m*g*cos

d1 = displacement along the incline = 2.6 m

= angle between Friction force and displacement = 180 deg, since given that block is pulled on the incline , and friction force is in opposite direction of motion

= angle of incline plane = 34 deg

Wf = *m*g*cos *d1*cos

Using given values:

Wf = 0.38*17*9.81*cos 34 deg*2.6*cos 180 deg

Wf = -136.6 J

Part D.

Work-done by gravity force up the incline will be:

Wg = Fg*d1*cos

Fg = Force of gravity along incline = m*g*sin

d1 = displacement along the incline = 2.6 m

= angle between gravity force and displacement = 180 deg, since given that block is pulled on the incline , and gravity force is in along the incline in downward direction

= angle of incline plane = 34 deg

Wg = m*g*sin *d1*cos

Using given values:

Wg = 17*9.81*sin 34 deg*2.6*cos 180 deg

Wg = -242.5 J

Part E.

Using Work-energy theorem for whole motion:

W_net = dKE

W1 + W2 + Wf + Wg = KEf - KEi

KEi = 0 since block starts from rest on horizontal ground

KEf = 0, since at the top of the incline block comes to rest

W1 = Work-done by tension force on the horizontal = 392.2 J (from part A)

Suppose block travels 'x' distance before coming to rest, then

W2 = Work-done by tension force along the incline = T*x*cos 0 deg = 83*x*cos 0 deg = 83x

Wf = Work-done by friction force along the incline = *m*g*cos *x*cos 180 deg = -*m*g*x*cos

Wg = Work-done by gravity force along the incline = m*g*sin *x*cos 180 deg = -m*g*x*sin

So, Using above values:

392.2 + 83*x - *m*g*x*cos - m*g*x*sin = 0 - 0

x = 392.2/[*m*g*cos +  m*g*sin - 83]

Using known values:

x = 392.2/[0.38*17*9.81*cos 34 deg + 17*9.81*sin 34 deg - 83]

x = 6.25 m = distance traveled along the incline before coming to rest

Please Upvote.