A force of 8.7 N is applied to a steel block initally at rest on a horizontal frictionless surface. The force, which is directed at an angle of 30.0o below the horizontal, gives the block a horizontal acceleration of 5.3 m/s2.
a) Draw a free body diagram and sum forces.
b) What is the normal force of the surface acting on the block?
Tcos@=ma .8.7Ncos30/5.3=1.4Kg.(b) w=mg=1.4times9.8=13.16
To solve this problem, we will break it down into two parts:
a) Drawing a free body diagram and summing forces:
First, let's draw a free body diagram for the steel block:
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Let's now identify the forces acting on the block. The applied force of 8.7 N directed at an angle of 30.0° below the horizontal can be broken down into its horizontal and vertical components. The vertical component (downward) does not affect the horizontal motion of the block, so we will only consider the horizontal component:
-----> F = 8.7 * cos(30°)
The normal force (N), which is exerted by the surface on the block, acts perpendicular to the surface and counterbalances the weight of the block. Since the block is on a frictionless horizontal surface, there is no horizontal friction force.
Therefore, the forces acting on the block are:
- Applied force (F) acting to the right.
- Normal force (N) acting upward.
To sum the forces, we need to use Newton's second law:
ΣF = ma
We know the mass of the block (which is not given in the problem) cancels out when solving for the normal force, so we can write:
N - F = 0
Rearranging the equation, we find:
N = F
So, in this case, the normal force (N) acting on the block is equal to the applied force (F), which is 8.7 N.
b) What is the normal force of the surface acting on the block?
As we determined above, the normal force (N) is equal to the applied force (F), which is 8.7 N.
Therefore, the normal force of the surface acting on the block is also 8.7 N.