Question

A 19.6 kg block is dragged over a rough, horizontal surface by a constant force of 148 N acting at an angle of 30.7 ◦ above the horizontal. The block is displaced 64.7 m, and the coefficient of kinetic friction is 0.185. 19.6 kg µ = 0.185 148 N 30.7 ◦ Find the work done by the 148 N force. The acceleration of gravity is 9.8 m/s 2 . Answer in units of J. 003 (part 2 of 2) 10.0 points Find the magnitude of the work done by the force of friction. Answer in units of J.

Answer 1

Answer:

W= +7224.65 J  : Work done by the 148 N force

$W_{Ff} = F_{f} * d= -35.53*64.7 = - 2299.1 J$ : Work done by the force of friction.

Explanation:

Work (W) is defined as the product of force F) by the distance (d)the body travels due to this force.

W= F*d Formula ( 1)

The work  is positive  (W+) if the force has the same direction of movement of the object.

The work  is negative (W-) if the force has the opposite direction of the movement of the object.

The component of the force that performs work must be parallel to the displacement.

Problem development

a)

Since the displacement of the block is horizontal, we calculate the horizontal component of the force (Fx):

F= 148 N ,  forming 30 degrees with the positive x axis

Fx= 148 N *cos30.7°

Fx= 111.66 N : Magnitude of the horizontal component of the force.

We apply formula (1)

W= (Fx)* (d) = 111.66 N * 64.7 m=7224.65 Joules = 7224.65 J

W= +7224.65 J : Work done by the 148 N force

b) We apply Newton's first law:

∑Fy = 0

∑F : algebraic sum of the forces in Newton (N)

N-W =0 Equation (1)

Where N is the normal force and W is the Weight of the block:

W= m*g = 19.6 kg * 9.8 m/s² = 192.08 N

We replace W =  192.08 N in the equation (1)

N - 192.08= 0

N = 192.08 N

Friction force : Ff

Ff= μ*N  Formula (2)

μ: coefficient of kinetic friction

N : Normal force (N)  

Data

µ = 0.185

N= 192.08 N

We replace data in formula (2)

Ff= μ*N

Ff= (0.185 )(192.08) N

Ff = 35.53N

Calculation of work done by frictional force:

We apply the formula (1) :

$W_{Ff} = F_{f} * d= -35.53*64.7 = - 2299.1 J$

Answer 2

Answer:

(a) The total work done by the 148 N force is 6838.92 J

(b) The magnitude of the work done by the force of friction is -1394.68 J

Explanation:

Here, we have, resolving the 148 N into vertical, Fv and horizontal, Fh components

Fv = 148 × sin 30.7 = 75.56 N

Fh = 148 × cos 30.7  = 127.26 N

Force of friction = Normal Force, F(N) × Coefficient of Friction, μ

Where:

Coefficient of Friction, μ = 0.185

Normal Force = 19.6 × 9.8 - 148 × sin 30.7 =  116.52 N

Force of friction = 116.52 N × 0.185 = 21.56 N Horizontal negative

Total Horizontal force required to pull = 127.26 N - 21.56 N = 105.7 N

The work done in pulling along the horizontal  is given by

Work done = Magnitude of Horizontal Force × Distance of force

Work done to pull = 105.7 N  × 64.7 m = 6838.92 J

While the work done less frictional work = 8233.6 J

(b) The magnitude of work done against friction =

Force of friction × Distance moved

= 21.56 N × 64.7 m = 1394.68 J

The magnitude of work done by frictional force = -1394.68 J.