Physics 2007 JAMB Past Questions - Question 49: A particle of weight 120 N is placed on a plane inclined at an angle 30 to the horizontal. If the plane has an efficiency of 60%, what is the force required to push the weight uniformly up the plane?

Correct Answer

B

Explanation

To determine the force required to push the weight uniformly up the inclined plane, you'll need to consider the forces acting on the particle. In this scenario, there are two main forces to consider: the gravitational force acting on the particle and the force required to overcome the friction on the inclined plane.

1. Gravitational Force (Weight):
The weight of the particle is acting vertically downward and can be calculated using the formula:

Weight (W) = Mass (m) x Acceleration due to gravity (g)

Given that the weight is 120 N, you can calculate the mass using:

m = W / g
m = 120 N / 9.8 m/s² ≈ 12.24 kg

2. Inclined Plane Forces:
The inclined plane creates two forces:

a. Normal Force (N): This force is perpendicular to the inclined plane and counteracts the vertical component of the weight. It can be calculated as:

N = W * cos(θ)

Where θ is the angle of the inclined plane, which is 30 degrees. Converting this to radians:

θ = 30 degrees * (π/180) ≈ 0.5236 radians

N = 120 N * cos(0.5236) ≈ 120 N * 0.866 ≈ 103.92 N

b. Frictional Force (F_friction):
The frictional force acts along the inclined plane and opposes the motion. It can be calculated as:

F_friction = µ * N

Where µ is the coefficient of friction. Since the efficiency of the plane is given as 60%, the effective force required to push the weight is 60% of the actual force required. Therefore:

Efficiency = (Useful Work Output / Total Work Input) = (F_required / W)

Efficiency = 0.60

Now, rearrange the equation to find F_required:

F_required = Efficiency * W
F_required = 0.60 * 120 N = 72 N

Now, calculate the frictional force:

F_friction = µ * N

To find µ, you can use the formula:

µ = F_friction / N

µ = 72 N / 103.92 N ≈ 0.692

Now, you have µ, and you can calculate the frictional force:

F_friction = µ * N
F_friction = 0.692 * 103.92 N ≈ 71.78 N

The force required to push the weight uniformly up the inclined plane is the sum of the weight component parallel to the plane and the frictional force:

F_required_total = Weight component parallel to the plane + Frictional force

F_required_total = W * sin(θ) + F_friction

F_required_total = 120 N * sin(0.5236) + 71.78 N

F_required_total ≈ 120 N * 0.866 + 71.78 N ≈ 103.92 N + 71.78 N ≈ 175.70 N

So, the force required to push the weight uniformly up the plane is approximately 175.70 N.