Answer
(a) 1600 N
(b) -4600 J
(c) 4600 J
(d) 0
Work Step by Step
Since the acceleration is zero, the force $F_p$ applied by the man must be equal in magnitude to the component of the piano's weight directed down the slope.
(a) $F_p = mg ~sin(\theta)$
$F_p = (380 ~kg)(9.80 ~m/s^2) ~sin(25^{\circ})$
$F_p = 1600~N$
(b) $Work = F_p ~d ~cos(180^{\circ})$
$Work = (1600 ~N)(2.9 ~m)(-1) = -4600 ~J$
(c) $Work = mg ~sin(\theta) \cdot d$
$Work = (380 ~kg)(9.80 ~m/s^2) ~sin(25^{\circ})(2.9 ~m)$
$Work = 4600 ~J$
(d) The net work done on the piano is 4600 J - 4600 J = 0
