What is the minimum work needed to push a 925-kg car 207 m up along a 14.5° incline? ignore friction.

4.7 x 10⁵ J

Explanation:

m = mass of the car = 925 kg

d = distance traveled parallel to incline = 207 m

θ = angle of the slope = 14.5 deg

Force applied to move the car up is given as

F = mg Sinθ                                               eq-1

minimum work needed is given as

W = F d

using eq-1

W =  mgd Sinθ

inserting the values

W = (925)(9.8)(207) Sin14.5

W = 4.7 x 10⁵ J

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west