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vagabundo [1.1K]

3 years ago

6

A car whose initial speed is 30 m/s slows uniformly to 10 m/s in 5 seconds. Determine the acceleration of the car. Sketch a gra

ph to support your answer. CHALLENGE: Determine the distance it travels in the 3 seconds.
This one is a little harder because you have to make a graph :(

1 answer:

Anton [14]3 years ago

5 0

Initial velocity=u=30m/s
Final velocity=v=10m/s
Time=5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{10-30}{5}$

$\\ \sf\longmapsto Acceleration=\dfrac{-20}{5}$

$\\ \sf\longmapsto Acceleration=-4m/s^2$

Now

For the second question

Time=3s

Using 2nd equation of motion

$\\ \sf\longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf\longmapsto s=30(3)+\dfrac{1}{2}(-4)(3)^2$

$\\ \sf\longmapsto s=90-2(9)$

$\\ \sf\longmapsto s=90-18$

$\\ \sf\longmapsto s=72m$

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Speed of the block at the bottom of the incline: 5.42 m/s

The first part of the problem can be solved by using the law of conservation of energy. Since the ramp is frictionless, the initial gravitational potential energy of the block at the top of the ramp is converted into kinetic energy at the bottom:

$mgh = \frac{1}{2}mv^2$ (1)

where

m is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

h is the initial height of the block

v is the speed of the block at the bottom

The initial height of the block is equal to the height of the ramp, so

$h=L sin \theta$ (2)

where

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$\theta=30^{\circ}$ is the angle of the ramp

Substituting (2) into (1) and re-arranging the equation, we find the speed

$2gL sin \theta = v^2$

$v=\sqrt{2gL sin \theta}=\sqrt{2(9.8)(3.00)sin 30^{\circ}}=5.42 m/s$

Coefficient of kinetic friction between the floor and the block: 0.3

In the second part of the motion, the block is slowed down by friction along the flat surface. According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the block:

$W=\Delta K=K_f -K_i$

where

W is the work done by friction

Kf is the final kinetic energy of the block, which is zero since the block comes to rest

$K_i = \frac{1}{2}mv^2$ is the initial kinetic energy of the block, where

m = 10.0 kg is the mass of the block

v = 5.42 m/s is its initial speed

Substituting into the equation, we find

$W=-\frac{1}{2}mv^2=-\frac{1}{2}(10.0)(5.42)^2=-146.9 J$

and the work is negative, since the direction of the force of friction is opposite to the direction of motion of the block.

Now we can rewrite the work as the product between the force of friction and the displacement of the block:

$W=-F_f d = - \mu mg d$

where

$\mu$ is the coefficient of friction

d = 5.00 m is the displacement of the block

Solving for $\mu$ ,

$\mu = - \frac{W}{mgd}=-\frac{-146.9}{(10.0)(9.8)(5.00)}=0.3$

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