Static friction 0.46 between a 0.2 kg block and a 2.1 kg cart. There is no kinetic friction between the cart and the horizontal surface. The acceleration of gravity is 9.8 m/s².

What minimum force F must be exerted on the 2.1 kg cart in order for the 0.2 kg block not to fall?

Static friction 046 between a 02 kg block and a 21 kg cart There is no kinetic friction between the cart and the horizontal surface The acceleration of gravity class=


Answer :

Answer:

49 N

Explanation:

According to Newton's second law of motion, the net force on an object is equal to its mass times its acceleration. By drawing a free body diagram, we can determine the forces on each mass and then apply Newton's law.

For the block, there are three forces:

Weight force mg pulling down,

Normal force N₁ pushing right,

Friction force N₁μ pushing up.

For the cart, there are five forces:

Weight force Mg pulling down,

Normal force N₂ pushing up,

Applied force F pushing to the right,

Normal force N₁ pushing left,

Friction force N₁μ pushing down.

Sum of forces on the block in the y direction:

∑F = ma

N₁μ − mg = 0

N₁μ = mg

Sum of forces on the block in the x direction:

∑F = ma

N₁ = ma

Substitute:

maμ = mg

aμ = g

a = g/μ

a = (9.8 m/s²) / (0.46)

a = 21.3 m/s²

Next, sum of forces on the cart in the x direction:

∑F = ma

F − N₁ = Ma

F − ma = Ma

F = (m + M) a

F = (0.2 kg + 2.1 kg) (21.3 m/s²)

F = 49 N