Respuesta :
Answer:
W= +7224.65 J : Work done by the 148 N force
[tex]W_{Ff} = F_{f} * d= -35.53*64.7 = - 2299.1 J[/tex] : 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) :
[tex]W_{Ff} = F_{f} * d= -35.53*64.7 = - 2299.1 J[/tex]
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.
