Free help with homework

Why join Brainly?

  • ask questions about your assignment
  • get answers with explanations
  • find similar questions

An object on mass 10kg moving with a velocity of 5.6 m/s on a rough horizontal surface comes to rest after traveling a distance of 16m. find

1. coefficient of friction
2. work done against friction
3. time taken by the body to come to rest


U= 5.6 m/s
v= 0.
mass= 10kg
s= 16m
2*a*s = v^2-u^2
2*a*16= (0)^2-(5.6)^2
32*a= -31.36
a= -31.36/32
a = -0.98 m/s^2
F = m*a
F = 10kg*-0.98m/s^2
F = -9.8 N
So,force of friction = 9.8 N
work done= F*S
= -9.8*16= -156.8J
Magnitude of negative work done is 156.8
To calculate time ,we will use first equation of motion,
v= u+a*t
0= 5.6+(-9.8)*t
-5.6 = -9.8*t
t = 5.6/9.8
t = 0.57142857
i.e. 6 seconds (approx.)
Hope this helped u
please mark as brainliest answer..........
1 5 1
while calculating the time, u took acceleration as -9.8 but earlier found it as -0.98    can u explain tht?
Srry it was by mistake,Now t = 5.71428571,and answer is still 6 seconds (approx.)
yeah, right. thnx again
ur welcome...
The Brain
  • The Brain
  • Helper
Not sure about the answer?
Learn more with Brainly!
Having trouble with your homework?
Get free help!
  • 80% of questions are answered in under 10 minutes
  • Answers come with explanations, so that you can learn
  • Answer quality is ensured by our experts