A. A 16.4 kg block is dragged over a rough, horizontal surface by a constant force of 114 N acting at an angle of [tex]31^\circ[/tex] above the horizontal. The block is displaced 42 m, and the coefficient of kinetic friction is 0.241. The acceleration of gravity is [tex]9.8 \, \text{m/s}^2[/tex].

Find the work done by the 114 N force. Answer in units of J.

B. Find the work done by the force of friction. Answer in units of J.

C. If the block was originally at rest, determine its final speed. Answer in units of m/s.

A. Work done by the applied force is 4490 J. B. Work done by the force of friction is -2912 J (negative because it acts in the opposite direction of motion). C. The final speed of the block is 3.12 m/s.

What is the net work done on the block?

The net work done on the block is the sum of the work done by the applied force and the work done by the force of friction, which is 4490 J - 2912 J = 1578 J.

A. The work done by the 114 N force can be found using the equation W = Fdcosθ, where F is the force, d is the displacement, and θ is the angle between the force and displacement. So,

W = (114 N)(42 m)cos(31°) = 4,206 J

Therefore, the work done by the 114 N force is 4,206 J.

B. The work done by the force of friction can be found using the equation W = fd, where f is the force of friction and d is the displacement. The force of friction can be found using the equation f = μFn, where μ is the coefficient of kinetic friction and Fn is the normal force.

Fn = mg = (16.4 kg)(9.8 m/s^2) = 160.72 N

f = μFn = (0.241)(160.72 N) = 38.76 N

So,

W = fd = (38.76 N)(42 m) = 1,628.32 J

Therefore, the work done by the force of friction is 1,628.32 J.

C. The net work done on the block is equal to the change in kinetic energy of the block, which is given by the equation ΔK = (1/2)mvf^2 - (1/2)mv0^2, where m is the mass of the block, vf is the final velocity, and v0 is the initial velocity (which is zero).

The net force on the block is the vector sum of the applied force and the force of friction, which can be found using the equation Fnet = Fapp - f. So,

Fnet = Fapp - f = (114 N) - (38.76 N) = 75.24 N

The acceleration of the block can be found using the equation a = Fnet/m. So,

a = Fnet/m = (75.24 N)/(16.4 kg) = 4.59 m/s^2

The final velocity of the block can be found using the equation vf^2 = v0^2 + 2ad, where d is the displacement. So,

vf^2 = (0 m/s)^2 + 2(4.59 m/s^2)(42 m) = 386.16

vf = sqrt(386.16) = 19.65 m/s

Therefore, the final speed of the block is 19.65 m/s.

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