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MrRissso [65]
2 years ago
6

John is pushing (horizontally) on a 100 kg (Fw = 980N) bench with an unknown force. If the coefficient of static friction is 0.2

0 and the acceleration is 1.84m/s/s, what is the applied force?
Physics
1 answer:
atroni [7]2 years ago
4 0

Answer:

1176N

Explanation:

Given parameters:

Mass of bench  = 100kg

Forward force  = 980N

Coefficient of static friction = 0.2

Acceleration  = 1.84m/s²

Unknown:

Magnitude of the applied force  =?

Solution:

Frictional force is a force that opposes motion, for a body to move, the applied force must be greater than the frictional force;

The Force applied;

  Force applied  = Frictional force + Weight of the body

   Frictional force  = umg

     u is the coefficient of static friction

      m is the mass

        g is the acceleration due to gravity

    Force applied  = 0.2 x 100 x 9.8 + 980 = 1176N

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A student pulls horizontally on a block with a spring scale. The block reads 24.5 Newtons before the block starts to move. It re
Ahat [919]

Answer:

The coefficient of friction causes the force on the object to be less than its initial reading on the spring scale.

Explanation:

Since the block reads 24.5 N before the block starts to move, this is its weight. Now, when the block starts to move at a constant velocity, it experiences a frictional force which is equal to the force with which the student pulls.

Now, since the velocity is constant so, there is no acceleration and thus, the net force is zero.

Let F = force applied and f = frictional force = μN = μW where μ = coefficient of friction and N = normal force. The normal force also equals the weight of the object W.

Now, since F - f = ma and a = 0 where a = acceleration and m = mass of block,

F - f = m(0) = 0

F - f = 0

F = f

Since the force applied equals the frictional force, we have that

F =  μW and F = 23.7 N and W = 24.5 N

So, 23.7 N = μ(24.5 N)

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Answer:

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