Answer:
10628.87 J
Explanation:
We are given that
Force applied =F=5592 N
[tex]\theta=30.1^{\circ}[/tex]
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=[tex]F\times displacement[/tex]
The perpendicular component of force=[tex]FSin\theta=5592sin(30.1)=2804.45N[/tex]
Work done =[tex]Fsin\theta\times D=2804.45\times 3.79=10628.87 J[/tex]
Hence, the work done in sliding the piano up the plank at a slow constant rate=10628.87 J
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.
