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Answer :
Final answer:
The work done by the 156 N force is 2630.77 J. The magnitude of the work done by the friction force is 743.14 J. The work done by the normal force is zero.
Explanation:
a. The work done by a force is calculated by multiplying the force applied by the displacement in the direction of the force. In this case, the force is applied at an angle of 31.9 degrees above the horizontal, so we need to find the component of the force in the horizontal direction.
Fx = F * cos(θ) = 156 N * cos(31.9°) = 132.3 N
The work done is given by W = Fx * d = 132.3 N * 19.9 m = 2630.77 J
b. The magnitude of the work done by the force of friction can be calculated using the formula:
Wfriction = friction force * displacement = μ * m * g * d, where μ is the coefficient of kinetic friction, m is the mass of the block, g is the acceleration due to gravity, and d is the displacement. Substituting the given values:
Wfriction = 0.209 * 18.8 kg * 9.8 m/s2 * 19.9 m = 743.14 J
c. The work done by the frictional force is negative, indicating that it acts against the motion of the block.
d. The work done by the normal force is zero because the displacement of the block is perpendicular to the direction of the normal force.
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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
