A 32 kg block of ice slides down a frictionless incline 2.4 m along the diagonal and 0.68 m high. A worker pushes up against the ice, parallel to the incline, so that the block slides down at constant speed. (a) Find the magnitude of the worker's force. How much work is done on the block by (b) the worker's force, (c) the gravitational force on the block, (d) the normal force on the block from the surface of the incline, and (e) the net force on the block

a) W = -213.26 J , b) T = 88.86 N , c)   W = 313.6 N , d)  N = 300.75 N , e♦1 F = 0

Explanation:

For this exercise we use Newton's second law, where the x axis is parallel to the plane and the y axis is perpendicular.

X axis

        T - Wₓ = 0

        T = Wₓ

Y Axis  

        N - Wy = 0

Let's use trigonometry to find the components

       sin θ = Wₓ / W

       cos θ = Wy / W

       Wx = W sin θ

       Wy = W cos θ

We substitute

     T = W sin θ

     N = W cos θ

Let's find the angle of the plane

     sin θ = y / L

     θ = sin⁻¹ 0.68 / 2.4

     θ = 16.46º

a) The work of the force of man (T)

        W = T L cos tea

         W = mg sin 16.46 L cos θ

         W = 32 9.8 2.4 sin 16.46 cos 180

         W = -213.26 J

As the block slides at constant speed, the work of the block is of equal magnitude, but the component of the weight is parallel to the displacement, so the angle is zeroº.

         W = 213.26 J

b) T = Wₓ

     T = mg sin θ

     T = 32 9.8 sin 16.46.

     T = 88.86 N

c) the gravitational force is the weight of the block

           W = mg

           W = 32 9.8

           W = 313.6 N

d) normal

         N = W cos 16.46

         N = 32 9.8 cos 16.46

        N = 300.75 N

e) as the block moves with constant speed the acceleration is zero therefore the net force is zero

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