A 10-kilogram block is pushed along a rough horizontal surface by a constant hor

Question

A 10-kilogram block is pushed along a rough horizontal surface by a constant horizontal force F. At time t=0, the velocity v of the block is 6.0 meters per secondin the same direction as the force. The coefficient of sliding friction is 0.2. Assume g=10 meters per second squared. a. calculate the force F necessary to keep the velocity constant. The force is now changed to a larger constant value F'. The block accelerates so that its kinetic energy increases by 60 joules while it slides a distance of 4.0 meters. b. calculate the force F'. c. Calculate the acceleration of the block.

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Explanation / Answer

a)
That For constant velocity
F = frictional force = µmg = 0.2*10*10 = 20N

b) In the
Net Force* distance = increase in energy.
(F’-F)*d = 60 => (F’-F) = 60/4 = 15 N
And F’= 15+20 = 35 N

C)
a = (F’-F)/m =15/10 = 1.5 m/s²
or
a = 60/(4*10) = 1.5 m/s²

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