Question

A 5592 N piano is to be pushed up a(n) 3.79 m frictionless plank that makes an angle of 30.1 ◦ with the horizontal. Calculate the work done in sliding the piano up the plank at a slow constant rate. Answer in units of J.

Answer 1

Answer:

10628.87 J

Explanation:

We are given that

Force applied =F=5592 N

$\theta=30.1^{\circ}$

Displacement=D=3.79 m

We have to find the work done in sliding the piano up the plank at a slow constant rate.

Work done=$F\times displacement$

The perpendicular component of force=$FSin\theta=5592sin(30.1)=2804.45N$

Work done =$Fsin\theta\times D=2804.45\times 3.79=10628.87 J$

Hence, the work done in sliding the piano up the plank at a slow constant rate=10628.87 J

Answer 2

Answer:s

W = - 10628.85 J

Explanation:

given,

weight of piano (mg)= 5592 N

distance of push = 3.79 m

angle = θ = 30.1°

displacement will in opposite direction of force (mg sinθ).

displacement make an angle of Φ i.e. 180° with the force

now,

Work done = F . s cos Φ

W = m g sin θ x s x cos (180°)

W = 5592 x sin 30.1° x 3.79 x cos (180°)

W = - 5592 x sin 30.1° x 3.79

W = - 10628.85 J

negative sign shows the direction of work done.