Question

A 5.70 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.4 N at an angle theta = 27.5° above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.10. What is the speed of the block 6.1 s after it starts moving?

I don't know how to do these equations in which they want you to find speed. A little help would be much appreciated.

Answer 1

This is a lot like your previous problem.

There is a force 10.4 cos 27.5 or 9.22 N horizontal from the cord.

There is a force 10.4 sin 27.5 or 4.80 N up from the cord

The force the block exerts down on the floor is the weight - the pull up from the cord
m g - 4.80 N
= 5.70(9.8) - 4.8 = 51.06 N normal to floor

So the friction force = .1 (51.06) = 5.11 N friction force
the net horizontal force on the block is therefore
9.22 - 5.11 = 4.11 N
now do F = m a
4.11 = 5.70 a
a = .721 m/s^2
initial speed o is 0 we assume
v = 0 + a t = .721 (6.1)= 4.4 m/s

Answer 2

Ohh ok great I understand.
Thanks a lot. :)