A 5.0-kg block at rest on a frictionless surface is acted
on by forces F1 = 5.5 N
and F2 = 3.5 N. What additional horizontal force will
keep the block at rest?
Answer: F3 = (-7.6 N)
Well, keeping a block at rest requires some serious work! So, let's break it down. The block is being acted on by the forces F1 and F2, which are 5.5 N and 3.5 N respectively. We need to find the additional horizontal force, let's call it F3, that will keep the block at rest.
Now, the net force on the block must be equal to zero for it to remain at rest. Since the surface is frictionless, we don't need to consider any opposing forces due to friction.
So, the equation to find the net force would be:
Fnet = F1 + F2 + F3
Now, let's plug in the given values:
0 = 5.5 N + 3.5 N + F3
Solving for F3, we have:
F3 = -7.6 N
So, it seems like the additional horizontal force required to keep the block at rest is a whopping -7.6 N. That's quite the force working against the other two! I hope that doesn't make things too chaotic!
To keep the block at rest, the sum of all the forces acting on it must be equal to zero. Therefore, the additional horizontal force, F3, can be calculated by finding the net force acting on the block.
The net force can be found by adding up all the forces acting on the block:
Net force = F1 + F2 + F3
Since F1 = 5.5 N and F2 = 3.5 N, we can substitute these values into the equation:
Net force = 5.5 N + 3.5 N + F3
Since the block is at rest, the net force must be equal to zero:
0 = 5.5 N + 3.5 N + F3
Simplifying the equation gives:
0 = 9 N + F3
To find the value of F3, we can isolate it:
F3 = -9 N
Therefore, the additional horizontal force needed to keep the block at rest is -9 N, or (-7.6 N) (rounded to one decimal place).
To find the additional horizontal force (F3) required to keep the block at rest, you need to determine the net force acting on the block. In this scenario, the net force should be zero since the block is at rest.
To calculate the net force, you need to sum up the forces acting on the block. The given forces are F1 = 5.5 N and F2 = 3.5 N. The net force can be calculated by adding the two forces together.
Net force (Fnet) = F1 + F2
Substituting the values, we get:
Fnet = 5.5 N + 3.5 N
= 9 N
Since the block is at rest, the net force acting on it should be zero. Therefore, an additional horizontal force is needed to balance out the existing forces and make the net force zero. This additional force is the negative sum of the existing forces:
F3 = -Fnet
= -9 N
So, the additional horizontal force required to keep the block at rest is F3 = -7.6 N.