Question of Physics

Question

A block of mass m is being pushed by a horizontal force F along

the inclined plane of inclination angle θ by distance of d along the plane (the block is inially at rest). At x = 0, a friction force starts acting on the block and the coecient of friction is given by µ = bx where b is a positive constant (Note that x axis is along the plane while y axis is normal to the plane). In terms of the given quantities (m, F, d, θ, b, g ), find

a) work done by external force F,

b) work done by the friction force, and

c) the speed of the block at x = d.

Answer 1

a) The work done by the horizontal force, F along the inclined plane is given by the dot product of that force vector and the distance vector traveled by the block, d. Since these vectors are parrel, The work done by the external force = Fd.

b) Taking the sum of differential work done by friction on the block across a distance, d, we have:

W = ∫₀^d F_f •dx = ∫₀^d F_f cos(180°)dx = ∫₀^d F_f (-1)dx = −∫₀^d F_f dx

where F_f is the force of friction as function distance = µF_n = (bx)(mg cos(θ)),

where F_n = (mg cos(θ) is the normal force of the block. If confused about this value lookup "inclined block problem" so that you can see a diagram, there are many examples.

Finally, we have:

W = −∫₀^d F_f dx = −∫₀^d (bx)(mg cos(θ)) dx = −(bmg)(cos(θ)) ∫₀^d x dx = −(bmg)(cos(θ)) ( (1⁄2)x² |^d₀ )

= −(bmg)(cos(θ)) ( (1⁄2)(d)² − (1⁄2)(0)²) = −((1⁄2)bmgd²)(cos(θ))

Answer 2

Part C: The force F must be strong enough to overcome the force of friction and the parallel component of the gravitational force. The portion of the force remaining after subtracting the frictional force and the gravitational component is what moves the mass up the slope. The work done by this net force is converted to kinetic energy at the top of the slope, 1/2mV². The kinetic energy can be equated to the difference of the total work and the sum of the frictional and gravitational work and solved for V.