Physics Help... Please..?

Question

A man pulls a block of mass m = 25 kg up an incline at a slow constant velocity for a distance of d = 5 m. The incline makes an angle q = 25° with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is µk = 0.4.

a) What is the work Wm done by the man?

At the top of the incline, the string breaks and the block, assumed to be at rest when the string breaks, slides down a distance d = 5 m before it reaches a frictionless horizontal surface. A spring is mounted horizontally on the frictionless surface with one end attached to a wall. The block hits the spring, compresses it a distance L = 0.8 m, then rebounds back from the spring, retraces its path along the horizontal surface, and climbs up the incline.

b) What is the speed v of the block when it first reaches the horizontal surface?

c) What is the spring constant k of the spring?

d) How far up the incline d1 does the block rebound?

Answer 1

a)
The work done = Force * Displacement
Force = Frictional force + Gravitational force
= m*µ*cos θ*g + m*g*sin θ
= 25*0.4*9.8*cos 25 + 25*9.8*sin 25
= 192.35 N
So work done = 192.35*5 =761.8 Joules

b)
Use the conservation of energy theorem.
The change in potential energy = mgd sin θ = 517.70 Joules
Work done against friction = µmg cosθ d = 0.4*25*9.8*cos 25 * 5 = 444.09 Joules

This is difference the increase in the kinetic energy = 517.70-444.09 =73.61 Joules

This is 0.5 mv2. So v = sqrt(2*73.61/25) = sqrt(5.888) = 2.42 m/s

c) The compression energy is equal to the kinetic energy at the bottom of the incline. So, we have
0.5 k x2 = 73.61
k = 5.888 N/m

d)
Again the kinetic energy is restored and it is used to combat the gravitational and frictional forces. The resultant of these forces is m*µ*cos θ*g + m*g*sin θ. This is 192.35 N (We have calculated it before)

So, the distance = Energy/Force = 73.61/192.35 = 0.38 m up the slope

Answer 2

a) this is the sum of the gain in potential energy
sin(25)*(m+M)*g*5
where m is the mass of the block and M is the mass of the man (assuming the man is also moving)

and the work done against friction
cos(25)*m*g*µk*5

b) the energy lost due to friction is the same
cos(25)*m*g*µk*5
the balance of potential energy os converted to kinetic energy of the block when it gets to the horizontal
sin(25)*m*g*5-
cos(25)*m*g*µk*5=
.5*m*v^2

To solve for v:
v=sqrt(2*g*5*(sin(25)-
cos(25)*µk))

c)
The kinetic energy gets converted to stored energy in the spring
.5*m*v^2=.5*k*x^2
since you computed v above, solve for k as
k=m*v^2/x^2
x is given
d) The kinetic energy .5*m*v^2 will be recovered from the spring and will be converted to potential energy less the frictional loss. Set d as the distance up the incline
.5*m*v^2=
sin(25)*m*g*d-cos(25)*m*g*µk*d

or
d=
v^2/(2*g*(sin(25)-cos(25)*µk)

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