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VARVARA [1.3K]
3 years ago
12

A custodian pushes a 119 kg box across a level floor. The box moves at a constant velocity. The custodian pushes horizontally on

the box with a force of 724 Newtons, what must be the friction force experienced by the box in Newtons?
Physics
1 answer:
bonufazy [111]3 years ago
8 0

The friction force experienced by the box in Newtons is 724N

The frictional force is the force that opposes the moving force acting on a body on a horizontal plane.

  • For a static body, the acceleration will be zero,

According to Newton's second law of motion:

  • ∑Fx = ma
  • +Fm - Ff = ma

Given the following paramters

  • Moving force Fm = 724N
  • acceleration a = 0m/s²
  • mass m = 119 kg

Substitute the given parameters into the formula to get the frictional force:

724 - Ff = 119(0)

724 - Ff = 0

Ff = 0 + 724

Ff = 724N

Hence the friction force experienced by the box in Newtons is 724N

Learn more here: brainly.com/question/8859573

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W = mg = 1700 (9.81) = 16677 N

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6 0
3 years ago
The momentum of a falling rock is found to be 200 kg m/s. What is the mass of the rock if it falls with a velocity of 5.0 m/s
Snezhnost [94]

Answer:

\boxed {\boxed {\sf 40 \ kilograms}}

Explanation:

Momentum is the product of velocity and mass. The formula is:

p=m*v

We know the rock is falling. Its momentum is 200 kilograms meters per second and its velocity is 5 meters per second. Substitute the values into the formula.

200 \ kg \ m/s = m * 5.0 \ m/s

We are solving for m, the mass. We must isolate the variable. It is being multiplied by 5 meters per second. The inverse of multiplication is division, so we divided both sides by 5.0 m/s.

\frac{200 \ kg \ m/s}{5.0 \ m/s}=\frac{ m* 5.0 \ m/s }{5.0 \ m/s}

\frac{200 \ kg \ m/s}{5.0 \ m/s}=m

The units of meters per second (m/s) cancel.

\frac{200 \ kg}{5.0 } =m

40 \ kg = m

The falling rock has a mass of <u>40 kilograms.</u>

4 0
3 years ago
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