You want to push a 71-kg box up a 16° ramp. The coefficient of kinetic friction between the ramp and the box is 0.23. With what magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?

345.60N

Explanation:

The magnitude of the force parallel to the ramp that will push the box up the ramp are the frictional force(Ff) and the moving force Fm

Total force = Ff + Fm

Ff = ηR

Ff = ηmgcosθ

Fm = Wsinθ

Fm = mgsinθ

Total force = ηmgcosθ + mgsinθ

Total force = mg(ηcosθ + sinθ)

Given

mass m = 71kg

acceleration due to gravity g = 9.8m/s²

angle of inclination θ = 16°

coefficient of friction η = 0.23

Substitute into the formula;

Total force = 71(9.8)(0.23cos16 + sin16)

Total force = 695.8(0.2211+0.2756)

Total force = 695.8(0.4967)

Total force = 345.60N

Hence the magnitude force parallel to the ramp that you should push on the box so that it moves up the ramp at a constant speed is 345.60N

what is the actual question though?

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2