A 20.0 N force is applied at an angle of 40.0 degrees above the horizontal to a 4.00 kg box. The box moves a horizontal distance of 4.00 meters. Friction is negligible. The work done by the 20.0 N force is
A) 75.0 J.
B) 61.3 J.
C) 50.1 J.
D) 46.3 J.
E) 40.5 J.

Question

Quizplus · Accepted Answer

The work done by a constant force is given by the equation:

W = Fdcosθ

where W is work, F is force, d is displacement, and θ is the angle between the force and the displacement vectors.

In this problem, F = 20.0 N, d = 4.00 meters, and θ = 40.0 degrees. We can find the horizontal and vertical components of the force using trigonometry:

Fhoriz = Fcosθ = (20.0 N)cos(40.0°) = 15.3 N
Fvert = Fsinθ = (20.0 N)sin(40.0°) = 12.9 N

Since the box moves horizontally, the work done by the vertical component of the force is zero. The work done by the horizontal component of the force is:

W = Fhoriz * d = (15.3 N)(4.00 m) = 61.2 J

Rounding to the nearest tenth gives:

W ≈ 61.3 J

Therefore, the best choice is B) 61.3 J.

A 200 N Force Is Applied at an Angle of 40