A block of mass 25 kg is pulled at a constant velocity along a rough horizontal

Question

A block of mass 25 kg is pulled at a constant velocity along a rough horizontal floor by an applied force of 50 N at an angle of 30 degree to the surface. What is the magnitude of the frictional force? What is the coefficient of friction between the block and the surface? The magnitude of the force (in newtons) required to cause a 3 kg object to move at 1.2 m/s in a circle of radius of 2 meters is: A 5 kg block is released from rest 75 meters above the ground. When it has fallen 50 meters, What is its kinetic energy? What is the potential energy of the object at this point? What is the total mechanical energy of the system?

Explanation / Answer

13.solution

by newtons third law, fnet=ma

acceleration in a uniform circle is ma=(v^2)/r

a = v2 / r*m

a = 1.22m/s / (2m*3kg)

a = 0.24 m/s2

fnet = ma = 3*0.24 = 0.72N

14.solution

(a) The acting gravity force on block is = g*sin theta.
Theta can be calculated as follows:
sin (50/75) = 0.011; therefore theta = 41.81 degrees
g sin 41.81 = 9.8*0.4 = 6.53 m/s^2
V final = V initial + a*t; V initial = 0.
V final = 6.53*t
Use the equation of displacement:
75 = 0.5*6.53*t^2
75 / 3.26 = 23 = t^2
t = 4.8 seconds
V final = 6.53*4.8=31.28m/s.

k.e = 0.5mV2 = 0.5*5*31.282

k.e = 2447.5 J

(b) potential energy = mgh = 5*9.8*25m

potential energy = 1225 J

(c) total mechanical energy = k.e + p.e = 3672.5 J

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