# M2 QuestionWatch

This discussion is closed.
#1

1) A car of mass 750kg moving along a level road, has its speed reduced from 25ms to 15ms by the brakes which produce a constant retarding force of 2250N. Calculate the distance covered whilst the speed is being reduced

2) The total mass of a cyclist and his machine is 140ks. The cyclist rides along a horizontal road against a constant total resistance of magnitude 50N. Find in J the total work done in increasing his speed from 6ms to 9ms whilst tracellinga distance of 30m.
0
13 years ago
#2
1)

a = F/m
a = -2250/750
a = -3 m/sΒ²

vΒ² = uΒ² + 2as
s = (vΒ² - uΒ²)/2a
s = 66.7m

or do it using energy:
work done in reduce speed = 0.5muΒ² - 0.5mvΒ²
W = 150,000 J

W = Fx
x = W/F
x = 66.7 m
0
#3
Thanks, so I've applied some of the methods you used there to the second question and in a round about way come up with the answere but I dont understand the theory behind it, could you explain what I've done?

Work done in increading speed = 0.5 x 140 x 81 - 0.5 x 140 x 36
= 3150

F= ma
a = 5/14

X - 3150 = Force x distance
= 1500
x= 4650 = answere in back of the book

So what does the x in the last equation stand for?
0
13 years ago
#4
2) The total mass of a cyclist and his machine is 140ks. The cyclist rides along a horizontal road against a constant total resistance of magnitude 50N. Find in J the total work done in increasing his speed from 6ms to 9ms whilst travelling a distance of 30m.

F - 50 = 140a
F is the resultant force forward, -50 is the resistive force. In the equation, F = ma, F is the resultant force.

u = 6m/s
v = 9m/s
a = ?
s = 30m

v^2 = u^2 + 2as
a = (v^2 - u^2)/2s = (9^2 - 6^2)/(2 x 30) = 0.75ms^-2

F - 50 = 140a
F = 140a + 50 = 140(0.75) + 50 = 155N

Work done = Fs = 155 x 30 = 4650Nm = 4650J

OR

Using energy considerations,
0.5mv^2 + mgh2 = a constant
0.5mv^2 + mgh2 = 0.5mu^2 + mgh1 + (F - R)s (F is the forward force and R is the resistive force)
0.5mv^2 + 0 = 0.5mu^2 + 0 + Fs - Rs
Fs = 0.5mv^2 - 0.5mu^2 + Rs

Fs = work done = 0.5(140)(9)^2 - 0.5(140)(6)^2 + 50(30)
Work done = 5670J - 2520J + 1500J = 4650J
0
#5
Thanks, but could you elaborate a little bit more on these lines:
0.5mv^2 + mgh2 = a constant
0.5mv^2 + mgh2 = 0.5mu^2 + mgh1 + (F - R)s
0
13 years ago
#6
(Original post by Bebop)
Thanks, but could you elaborate a little bit more on these lines:
0.5mv^2 + mgh2 = a constant
0.5mv^2 + mgh2 = 0.5mu^2 + mgh1 + (F - R)s
Right, this is the work-energy principle (page 69 edexcel M2 book)
0.5mv^2 + mgh2 = the final energy of the cyclist at the end of its journey.
=> 0.5mv^2 + mgh2 = 0.5mu^2 + mgh1 (as long as there are no external forces on the body)
- 0.5mv^2 is the final KE, mgh2 is the final GPE.
- Since the cyclist is travelling horizontally, the initial and final GPE = 0.
- But because there are external forces on the bicycle, we have to include them.

so;
0.5mv^2 = 0.5mu^2 + (F - R)s
0.5mv^2 - 0.5mu^2 = (F - R)s

which is correct because;
gain in KE = work done by the cyclist.

If there was no resistive force, then the equation would be
0.5mv^2 - 0.5mu^2 = Fs

but because there is a resistive force, working against the cyclist then it is negative.
hence the equation;
0.5mv^2 - 0.5mu^2 = Fs - Rs = (F - R)s
0
#7
Brilliant!

Cheers that makes sence.
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