Question

A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 8.0 m/s2. A 20 kg mass slides along the top of the 100 kg mass and has an acceleration of 3.0 m/s2. (It thus slides backward relative to the 100 kg mass.) (a) What is the frictional force exerted by the 100 kg mass on the 20 kg mass? (b) What is the net force acting on the 100 kg mass?

Answer 1

a) 60 N

b) 860 N

Step-by-step explanation:

Given that,

`m_1` = 100 kg

`m_2` = 20 kg

`a_(1)` = 8.0
`(m)/(s^(2))`

`a_(2)` = 3.0
`(m)/(s^(2))`

a) By Newton's Law,

∑
`F_(m_2,x) = f_k = m_(2) * a_(2)`

∑
`F_(m_2,y) = F_(N) - m_(2) * g = 0`

Hence,

`f_k = m_2 * a_2 = 20 * 3 = 60 N`

b) By Newton's Law

∑
`F_(m_1,x) = F = m_(1) * a_(1)`

Hence,

`F = m_(1) * a_(1) = 100 * 8 = 800 N`

Net force acting on 100 kg mass,

`F_(net) = F + f_k = 800 + 60 = 860 N`