Starting from rest, a student pulls a (34) kg box a distance of (13.50) m across the floor using a (161) N force applied at 35.0 degrees above horizontal. The coefficient of kinetic friction is 0.220 between the floor and the box. Find the work done by the pulling force. Give your answer in joules (J) and with 3 significant figures.
work = pulling force in direction of motion * distance
like period, the friction and all are irrelevant.
work = 161 cos 35 * 13.5
If there are more parts like how fast does it go or something, the rest matters, but that is all you need to know for the question you asked.
To find the work done by the pulling force, we can use the equation:
Work = Force × Displacement × cos(θ)
Where:
Force is the magnitude of the pulling force,
Displacement is the distance the box is pulled, and
θ is the angle between the force and the direction of displacement.
Given:
Mass of the box (m) = 34 kg
Distance pulled (d) = 13.50 m
Force (F) = 161 N
Angle (θ) = 35.0 degrees
Coefficient of kinetic friction (μ) = 0.220
First, let's calculate the magnitude of the normal force (Fn) acting on the box. This is the force exerted by the floor on the box and is equal in magnitude but opposite in direction to the force due to gravity on the box.
Fn = mg
Where:
m is the mass of the box
g is the acceleration due to gravity (approximately 9.8 m/s²)
Fn = (34 kg) × (9.8 m/s²)
Fn ≈ 333.2 N
Next, let's calculate the magnitude of the frictional force (Ff) acting on the box. This can be found using the equation:
Ff = μ × Fn
Ff = (0.220) × (333.2 N)
Ff ≈ 73.1 N
Since the box is moving, the direction of kinetic friction is opposite to the direction of the pulling force. Therefore, the magnitude of the net force acting on the box is:
Net force = Force - Frictional force
Net force = 161 N - 73.1 N
Net force ≈ 87.9 N
Next, let's calculate the horizontal component of the pulling force (Fhx) using the angle θ:
Fhx = F × cos(θ)
Fhx = (161 N) × cos(35.0 degrees)
Fhx ≈ 132.0 N
The work done by the pulling force is then given by:
Work = Fhx × d
Work = (132.0 N) × (13.50 m)
Work ≈ 1782 J
Therefore, the work done by the pulling force is approximately 1782 joules (J).
To find the work done by the pulling force, we need to calculate the net force acting on the box and multiply it by the displacement.
First, let's calculate the gravitational force acting on the box. The formula for gravitational force is:
F_gravity = m * g
where m is the mass of the box (34 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Substituting the values:
F_gravity = 34 kg * 9.8 m/s^2 = 333.2 N
Next, we need to find the frictional force. The formula for frictional force is:
F_friction = u * F_normal
where u is the coefficient of kinetic friction (0.220) and F_normal is the normal force. The normal force is equal to the gravitational force on the box, which we just calculated. Substituting the values:
F_friction = 0.220 * 333.2 N = 73.104 N
Now, let's find the net force acting on the box. The net force is the vector sum of the applied force and the frictional force. Since the angle between the applied force and the displacement is given as 35 degrees above the horizontal, we need to calculate the horizontal component of the applied force. The horizontal component is given by:
F_horizontal = F_applied * cos(theta)
where F_applied is the applied force (161 N) and theta is the angle (35 degrees). Substituting the values:
F_horizontal = 161 N * cos(35 degrees) = 131.27 N
The net force is then:
F_net = F_horizontal - F_friction = 131.27 N - 73.104 N = 58.166 N
Finally, we can calculate the work done by the pulling force using the formula:
Work = F_net * d
where d is the displacement (13.50 m). Substituting the values:
Work = 58.166 N * 13.50 m = 784.932 J
Therefore, the work done by the pulling force is approximately 785 J (to 3 significant figures).