Two blocks, M = 14.2 kg and m = 4.73 kg, are connected via a massless rope. They are being pushed up a frictionless hill, with a slope of 15.80, with a constant force in the direction of the incline, such that there is a total acceleration of 2.53 m/s2 for the system

Slope is rise over run, meaning the ratio of the change in height to the change in horizontal distance. So a slope of 15.80 corresponds to an angle of ascension θ of

tan(θ) = 15.80   →   θ ≈ 86.38°

The order of the blocks (i.e. whether the large one is pulling the smaller one up or vice versa) does not matter, since friction is not a concern. So if we take the connected blocks as a single mass, by Newton's law we have a net force acting parallel to the incline of

∑ F = P - (M + m) g sin(θ) = (M + m) a

(see the attached free body diagram)

where

P = magnitude of the push

g = 9.80 m/s²

a = 2.53 m/s²

and M and m are the given masses.

Then the system requires a push of

P = (M + m) (a + g sin(θ))

P = (14.2 kg + 4.73 kg) (2.53 m/s² + (9.80 m/s²) sin(86.38°))

P ≈ 233 N

If you have to find the tension in the rope, consider the free body diagram for one of the blocks. By Newton's second law, the net parallel force acting on, say, the larger block (if it's being pulled by the rope) is

∑ F = T - M g sin(θ) = M a

where

T = tension in the rope

Then

T = M (a + g sin(θ))

T = (14.2 kg) (2.53 m/s² + (9.80 m/s²) sin(86.38°))

T ≈ 175 N

answer: good job it's all correct !

explanation:

i think the answer would be b