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snow_tiger [21]

3 years ago

14

An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov

e the horizontal. The block is displaced by 19.9 m, and the coefficient of kinetic friction is 0.209.
a. Find the work done by the 156 N force. The acceleration of gravity is 9.8 m/s^2
b. Find the magnitude of the work done by the orce of friction
c. What is the sign of the work done by the frictional force?
d. Find the work done by the normal force

1 answer:

Sav [38]3 years ago

4 0

Answer:

a) W = 2635.56 J

b) Wf = 423.27 J

c) c) The Sign of the work done by the frictional force (Wf) is negative (-)

d) W=0

Explanation:

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

The formula for calculate the work is :

W = F*d*cosα

Where:

W : work in Joules (J)

F : force in Newtons (N)

d: displacement in meters (m)

α :angle that form the force (F) and displacement (d)

Known data

m = 18.8 kg : mass of the block

F= 156 N,acting at an angle θ = 31.9◦°: angle above the horizontal

μk= 0.209 : coefficient of kinetic friction between the cart and the surface

g = 9.8 m/s²: acceleration due to gravity

d = 19.9 m : displacement of the block

Forces acting on the block

We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.

W: Weight of the cart : In vertical direction downaward

N : Normal force : In vertical direction the upaward

F : Force applied to the block

f : Friction force: In horizontal direction

Calculated of the weight of the block

W= m*g = ( 18.8 kg)*(9.8 m/s²)= 184.24 N

x-y components of the force F

Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N

Fy = Fsinθ = 156 N*sin(31.9)° = 82.44 n

Calculated of the Normal force

Newton's second law for the block in y direction :

∑Fy = m*ay ay = 0

N-W+Fy= 0

N-184.24+82,44= 0

N = 184.24-82,44

N = 101.8 N

Calculated of the kinetic friction force (fk):

fk = μk*N = (0.209)*( 101.8)

fk = 21.27 N

a) Work done by the F=156N.

W = (Fx) *d *cosα

W = (132.44 )*(19.9)(cos0°) (N*m)

W = 2635.56 J

b) Work done by the force of friction

Wf = (fk) *d *cos(180°)

Wf = (21.27 )*(19.9) (-1) (N*m)

Wf = - 423.27 J

Wf = 423.27 J :magnitude

c) The Sign of the work done by the frictional force is negative (-)

d) Work done by the Normal force

W = (N) *d *cos(90°)

W = (101.8 )*(19.9) (0) (N*m)

W = 0

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