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Please help.: Sliding from left to right in a straight line on a horizontal steel surface, an aluminum block weighing 20 newtons is acted on by a 2.4 newton friction force. The block will be brought to rest by the friction force in a distance of 10 meters. Determine the magnitude of the acceleration of the block as it is brought to rest by the friction force. (Show all work)

Answer :

syed514
Mass of the block = 2/g = 2/9.8 =0.204 kg. 

Frictional force = mass * acceleration 

a = F/ m where F is the frictional force = 2.4 

a = 2.4/ 0.204 = 11.76 m/s^2 

Answer:

-1.176 m/s²

Explanation:

Given:

weight of aluminium block is w = 20.0 N

frictional force, f = 2.4 N

final velocity, v = 0

distance covered, d = 10 m

From Newton's second equation of motion,

F = ma

where, m is the mass and a is the acceleration.

mass of the block, m g = 20.0 N

m = 2.04 kg

The block is brought to rest by friction force:

f =- m a

⇒ 2.4 N = -(2.04 kg)(a)

⇒a = -1.176 m/s²

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