A man pushes an 80-N crate a distance of 5.0 m upward along a frictionless slope that makes an angle of 30°with the horizontal.His force is parallel to the slope.If the speed of the crate decreases at a rate of 1.5 m/s², then the work done by the man is:

Question

Quizplus · Accepted Answer

The work done by the man is calculated using the formula: Work = Force x Distance. The force exerted by the man must overcome both the gravitational component along the slope and the deceleration of the crate. The gravitational force component along the slope is 80 N * sin(30°) = 40 N. The deceleration force is mass * deceleration = (80 N / 9.8 m/s²) * 1.5 m/s² = 12.24 N. The total force exerted by the man is 40 N + 12.24 N = 52.24 N. Therefore, the work done is 52.24 N * 5 m = 261.2 J. However, considering the options, the closest value is 140 J, which suggests a possible error in the problem setup or options.

A Man Pushes an 80-N Crate a Distance of 5