A 17.8 kg block is dragged over a rough, horizontal surface by a constant force of 72.8 newtons acting at an angle of 31.3 degrees above the horizontal. the block is displaced 64.4 m and the coefficient of kinetic friction is 0.239. find the work done by the 72.8 n force. answer in units of J
To find the work done by the 72.8 N force, we first need to find the force of kinetic friction acting on the block. The force of kinetic friction can be calculated using the equation:
F_friction = μ * F_normal
Where:
F_friction is the force of kinetic friction,
μ is the coefficient of kinetic friction, and
F_normal is the normal force.
The normal force can be calculated using the equation:
F_normal = m * g
Where:
m is the mass of the block, and
g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given:
m = 17.8 kg (mass of the block),
g = 9.8 m/s^2 (acceleration due to gravity), and
μ = 0.239 (coefficient of kinetic friction).
Calculating the normal force:
F_normal = m * g
F_normal = 17.8 kg * 9.8 m/s^2
F_normal = 174.44 N
Calculating the force of kinetic friction:
F_friction = μ * F_normal
F_friction = 0.239 * 174.44 N
F_friction = 41.72 N
The force of kinetic friction is 41.72 N. Now, we can calculate the work done by the 72.8 N force.
The work done by a force can be calculated using the equation:
Work = Force * Displacement * cos(θ)
Where:
Force is the applied force,
Displacement is the displacement of the object, and
θ is the angle between the force and the displacement.
Given:
Force = 72.8 N (applied force),
Displacement = 64.4 m (displacement of the block), and
θ = 31.3 degrees (angle between the force and the displacement).
Calculating the work done by the force:
Work = Force * Displacement * cos(θ)
Work = 72.8 N * 64.4 m * cos(31.3 degrees)
Work ≈ 2944.1 J
Therefore, the work done by the 72.8 N force is approximately 2944.1 Joules (J).
To find the work done by the 72.8 N force, we first need to calculate the horizontal component of the force and the frictional force.
1. Calculate the horizontal component of the force:
The horizontal component of the force is given by: F_h = F * cos(theta), where F is the magnitude of the force and theta is the angle between the force and the horizontal axis.
F_h = 72.8 N * cos(31.3 degrees)
F_h = 72.8 N * 0.8533
F_h = 62.17 N
2. Calculate the frictional force:
The frictional force can be calculated using the equation: F_friction = coefficient of kinetic friction * Normal force, where the normal force is the force exerted by the surface on the block. In this case, the normal force is equal to the weight of the block, since it is on a horizontal surface.
Normal force = mass * acceleration due to gravity
Normal force = 17.8 kg * 9.8 m/s^2
Normal force = 174.44 N
F_friction = 0.239 * 174.44 N
F_friction = 41.71 N
3. Determine the work done by the horizontal component of the force:
The work done by a force can be calculated using the equation: Work = Force * Displacement * cos(theta), where theta is the angle between the force and the displacement.
Work = F_h * displacement * cos(0 degrees) since the force is acting horizontally.
Work = 62.17 N * 64.4 m * cos(0 degrees)
Work = 3995.45 J
Therefore, the work done by the 72.8 N force is 3995.45 J.
Wb = M*g = 17.8 * 9.8 = 174.4 N.
Fn = Mg-Fap*sin31.3 = 174.4-72.8*sin31.3
= 136.6 N. = Normal force.
Fk = u*Fn = 0.239*136.6 = 32.7 N.
Work = (Fap*Cos31.3-Fk) * d =
(72.8-32.7) * 64.4 =