# 1)a body of mass 5 kg is under the action of 50N on the horizontal surface .if the coefficient of friction in bet the surface is 1 the distance it travels in 3 seconds is 2)a block of weight 100N is pushed by a force F on a horizontal rough plane moves with an acceleration 1m/s2^ the coefficient of friction is 3) a horizontal force applied on a body on a rough horizontal surface produces an acceleration A,if coefficient of friction bet body and surface which is m is reduced to m/3,the acceleration increases by 2 units.the value of m is PLZZZZZ PUT THE QUE NO. BEFORE ANSWERING

1
by aimanakhtar799

2014-10-05T10:05:53+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
1)

Frictional force on the body = μ m g = 1 * 5 kg * g = 5 g   N
Net force acting on the body = 50 N - friction force = 50 N - 5 g

If g = 10 m/s², then net force = 50 - 50 N = 0. So acceleration is 0. Velocity = 0.
The body stays at rest.

If g = 9.81 m/s² then net force = 50 - 5 * 9.8 = 1 N
acceleration = net force / mass = 1 / 5 = 0.2 m/sec²

Distance traveled in 3 seconds is S = u t + 1/2 a t² = 0 + 1/2 * 0.2 * 3² = 0.9 meters

2)

Net force = mass * acceleration = 100N/ g * 1 m/s² = F - μ (m g) = F - μ 100 N

μ =  [ F - 100 / g ] / 100

If g = 10 m/s²,        μ = (F - 10)/100
If g = 9.81 m/s²,     μ = ( F - 10.19)/100

3)

m A = net force = Force applied F - friction μ m g
A = (F - μ m g)/m = F/m - μ g   --- equation 1

acceleration increases by 2 units, if μ is reduced to 1/3μ.
A + 2 = F/m - (μ/3) g        ---  equation 2

( F/m - μ g)  + 2 =  F/m - μ g /3
2 μ g /3 = 2

μ = 3/g     = 0.3  if g =10m/s²   or   0.306,   if g = 9.81 m/s²

select best answ
i would provided anyone dared to answer it
thanx