Question

A block of mass 5.4 kg is pulled by a force of 76.79 N. As a result, it moves to the right. The coefficient of kinetic friction is 0.11. whats the acceleration…. PLEASE HELP

Answer 1

Newton's second law of motion: sum of forces = mass*acceleration. That's a good start. Then we have to think about the forces. In the x-direction, there are a frictional force R, and the force F pulling the block. Let the positive direction be in favour of F. Then the sum of forces in x-direction is \[F - R = ma\]

Answer 2

Are you able to solve it from there, or do you need help how to find the frictional force R?

Answer 3

Hint: \[R =\mu*N\], where N is the normal force and \[\mu\] is the coefficient of kinetic friction.

Since the block is pulled only in the x-direction, the net forces in y-direction is 0 (normal force - weight = 0). Hence, normal force = weight, N = G, where G is the weight of the block of mass. That implies that \[R = \mu*N = \mu*G = \mu*m*g = 0.11*5.4*9.81\] Newtons.

From here on, just put \[R\] into Newton's second law, and solve for a :)

Answer 4

That was a large hint indeed! :D

Answer 5

hmmmm let me see if i can do it I'm really not good at this stuff but ill try it
thank you soooo much

Answer 6

ok no i don't get it…..???

Answer 7

The last line will be:
\[F -R = ma => a = \frac{ F-R }{ m } = \frac{ 76.79 Newtons - 0.11*5.4kg*9.81m/s^2}{ 5.4 kg } \approx 13.1 Newtons.\]

Answer 8

*13.1 Newtons

Answer 9

THANKS SO MUCH!!!!

Answer 10

No problem :)