Thinking about the winter we missed out on this year. Calvin and his tiger go sledding down a snowy hill. There is friction between the snow and the sled, and there is air resistance. Looking at the entire time that they spend on the hill (from when they start from rest at the top of the hill until they reach the bottom moving fast), indicate whether each of the following quantities is positive, negative, or zero, by writing either, -, or 0 for the questions below. We take Calvin, the tiger, and the sled as our system.
a. The work done by the normal force that the snow exerts on the system?
b. The work is done by the frictional force that the snow exerts on the system?
c. The change in the kinetic energy of the system?
d. The change in the gravitational potential energy of the system?
e. The total work is done on the system?

a)  W=0, b) Work is negative, c) work is positive and scientific energy variation is positive, d)     the variation of the potential enrgy is negative,

e) total work is positive

Explanation:

Work in physics is defined by the scalar scalar product of force by displacement

          W = F. dx

The bold are vectors; this can be written in the form of the mules of the quantities

          W = F dx cos θ

where θ is the angle between force and displacement.

a) The normal force is perpendicular to the inclined plane which is perpendicular to the displacement, therefore the angle is

         θ = 90         cos 90 = 0

        W=0

In conclusion the work is zero

b) The friction force opposes the displacement whereby the angle is

       θ = 180      cos 190 = -1

        W = - fr d

Work is negative

c) To calculate the change in kinetic energy we use that the work is equal to the variation of the kinetic energy

            m g sin θ  L = ΔK

this magnitude is positive since the angle is zero cos 0 = 1

how the system starts from rest ΔK = Kf -K₀=  + Kf -0

work is positive and scientific energy variation is positive

d) change in potential energy

               The potential energy is is ΔU = Uf -U₀

we fix the reference system in the bases of the plane so Uf = 0

               ΔU = -U₀

         the variation of the potential enrgy is negative

e) The total work is formed by the work of the weight component, the work of the friction force

              W_Total = W_weight - W_roce

as the body moves down

              W_Total> 0

Therefore the total work is positive

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.