A rightward force of 246 N is applied to a 35 N suitcase to accelerate it across the floor. The coefficient of friction between

Question

anygoal [31]

3 years ago

10

A rightward force of 246 N is applied to a 35 N suitcase to accelerate it across the floor. The coefficient of friction between

the crate and the floor is 0.75. What is the acceleration of the crate?

Physics

1 answer:

katrin [286] · Answer 1 · 2022-09-01T23:20:02+03:00

katrin [286]3 years ago

5 0

(35/g) = mass of 3.57kg. (9.8 for g).
 (35 x 0.75) = 26.25N. friction.
 (246 - 26.25) = net 219.76N. accelerating force.
 Aceleration = (f/m) = (219.76/3.57) = 61.56m/sec^2