Question

Part 1) A 1.0 kg block is pushed 2.0 m at a constant velocity up a vertical wall by a constant force applied at an angle of 29.0 ◦ with the horizontal, as shown in the figure. The acceleration of gravity is 9.81 m/s 2. Drawing not to scale. If the coefficient of kinetic friction between the block and the wall is 0.20, find a) the work done by the force on the block. Answer in units of J.

Part 2) the work done by gravity on the block. Answer in units of J.

Part 3) The magnitude of the normal force between the block and the wall.

Answer 1

let m = 1.0 kg
d = 2.0 m
theta = 29 degrees.
let F is the applied force.
let Fx and Fy are components of F.
As the block is moving upward with constant velocity, net force acting on the block must be zero.

Fnety = 0

Fy - m*g = 0

Fy = m*g

= 1*9.81

= 9.81 N

a) The workdone by the force on the block, WF = Fy*d*cos(0)

= 9.81*2

= 19.62 J <<<<<<<<<<-----------------Answer

b) The workdone by gravity, Wg = Fg*d*cos(180)

= 1*9.81*2*(-1)

= -19.62 J <<<<<<<<<<-----------------Answer

c) we know, Fy = F*sin(29)
Fx = F*cos(29)

Fy/Fx = tan(29)

Fx = Fy/tan(29)

= 9.81/tan(29)

= 17.7 N

now use, Fnetx = 0

Fx - N = 0

N = Fx

= 17.7 N <<<<<<<<<<-----------------Answer