I tried EVERYTHING to solve this problem but no luck. It concerns newton's second law. i know it's easy. help.

Question

A 4.04 kg block located on a horizontal floor is pulled by a cord that exerts a force F = 10.3 N at an angle q = 16.0° above the horizontal. The coefficient of kinetic friction between the block and the floor is 0.070. What is the speed of the block 3.70 s after it starts moving?
I know that static friction can be ignored. Any idea how to solve this? I worked on it for over an hour. Please help.
Thank you....

Answer 1

ID all the forces:
tension (T) on the cord, weight (W), kinetic friction (K = kN).

Add them up in the vertical and horizontal directions:

Vertical:
weight - vertical pull on the cord = W - T sin(16deg) = mg - T sin(16deg); where m = 4.04 kg and g = 9.81 m/sec^2 T = F = 10.3 Newton.

Horizontal:
horizontal pull - kinetic friction = T cos(16deg) - kN = T cos(16deg) - k(W - T sin(16deg)) = net horizontal force f = ma; where N = normal force on the block = W - vertical pull and k = .07 friction coefficient. [NB: The normal force (N) is not the weight of the block because there is a vertical pull by the cord that offsets the weight.]

Solve the horizontal forces for a = [T cos(16deg) - k(mg - T sin(16deg))]/m = f/m

With a determined, solve v = at; where t = 3.7 sec. v is the answer you are looking for.

Everything's set up, you can do the math. But the important thing is to learn the steps to solving net force problems like this. First, ID all the forces. There will be acceleration if and only if the net forces are <> 0. One the other hand, if all the forces add up to zero, there will be no acceleration as f = 0 = ma implies a = 0.

Then write all the vertical forces in one equation and all the horizontal ones in another. If the moving object of mass m is accelerating horizontally, then set the horizontal equation = fx = max <> 0. If there is no vertical acceleration, set that equation = fy = may = 0.

If it's accelerating vertically set the vertical forces equation = fy = may <> 0. If there is no horizontal acceleration, then the horizontal forces equation = fx = max = 0.

Answer 2

Resolve the force F into vertical and horizontal forces.
Horizontal force = F cos θ = 10.3 cos 16 = 9.9 N.

Vertical upward force = F sin θ = 10.3 sin 16 = 2.84 N.
mg is the weight acting down. But this is reduced by 2.84 N.
Hence the net down ward force is 4.04 *9.8 - 2.84 = 36.752 N

Hence the normal reaction acting upward is 36.752 N

The kinetic friction acting horizontally opposing the motion is
0.07 x normal reaction = 0.07 x 36.752 N = 2.57 N.
The Net pulling force horizontally is F cos θ - μ [normal reaction]
= 9.9 - 2.57 = 7.33 N

Hence the block is accelerated from zero to some value.
Acceleration = Force / mass = 7.33 / 4.04 = 1.82 m/s^2.
Velocity in time t = a t [since u = 0]
= 1.82 * 3.7 = 6.734 m/s.
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Answer 3

enable F be the stress sought. you will agree that internet stress on the middle link will reason the decrease 2 links to boost up on the fee you already got here across. it fairly is, ma = F ? w = F ? mg. as a result, F = m(a + g), the place m is the mass sped up by way of stress F, i.e., the mass of the two decrease links, amounting to 0.seventy 4 kg. as a effect, F = 0.seventy 4 (2.80 one + 9.8) = 9.33 N. by way of the comparable token, the stress the middle link has to exert on the decrease link is F' = 0.37 ( 2.80 one + 9.8) = 4.sixty seven N. If entire chain mass is used, the ensuing stress is 14 N, because it is going to. elementary.