A 15kg box is pushed across the floor with an acceleration of 3 m/s2 with an applied force of 65N. What is the normal force, the net force, the force of friction, and the coefficient of friction between the box and the ground?

0.034

Explanation:

a) Friction opposes motion, so is in the opposite direction of the net force on the sled. Taking "left" to be "positive", the two animals are applying a net force of +42 N -15 N = 27 N. That is to the left, so the friction force is to the right.

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b) The total of forces applied by the animals and friction is ...

 27 N -5 N = 22 N . . . to the left

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c) Acceleration can be found from the relation ...

 a = F/m . . . . 'm' is mass

 a = (22 N)/(15 kg) = 22/15 m/s² . . . . 'm' is meters

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d) The coefficient of friction is the ratio of friction force to the normal force on the surface producing the friction. Here, that normal force is the force due to gravity acting on the 15 kg mass of the sled.

coefficient of friction = (5 N)/(9.8 m/s²·15 kg) = 5/147

coefficient of friction ≈ 0.034

your answer would be fusion

f=ma newton's second law 7x8=56n