A constant force of 45 N directed at angle $\theta$ to the horizontal pulls a crate of weight 100 N from one end of a room to another a distance of 4.0 m. Given that the vertical component of the pulling force is 12 N, calculate the work done by the force in moving the crate. A) 4.1 x 10² J B) 4.8 x 10¹ J C) 3.9 x 10² N/m D) 1.7 x 10² J E) 3.9 x 10³ J

Question

Quizplus · Accepted Answer

The horizontal component of the force is calculated using $ F_x = \sqrt{F^2 - F_y^2} = \sqrt{45^2 - 12^2} = 43.6 \, \text{N} $. The work done is $ W = F_x \cdot d = 43.6 \, \text{N} \times 4.0 \, \text{m} = 174.4 \, \text{J} \approx 1.7 \times 10^2 \, \text{J} $.

A Constant Force of 45 N Directed at Angle θ\thetaθ To the Horizontal Pulls a Crate of Weight 100 N

A Constant Force of 45 N Directed at Angle $\theta$ To the Horizontal Pulls a Crate of Weight 100 N