Question

4. A child of mass 30 kg skates down from the top of a ramp having a constant slope of
15. The child's speed increases from 1.5 ms to 3.0ms' as she reaches the
bottom of the ramp. A force of kinetic friction of magnitude 50 N opposes her
motion. Draw the appropriate force diagram and determine the length of the ramp.
Take g = 10ms-2​

Answer 1

Answer:

I dont know that much but i have valuable information that when something falls its speed increases by 9m/s per 1 second.

Answer 2

The length of the ramp is 3.66 m

Explanation:

Given that,

Mass = 30 kg

Angle = 15 °

Force = 50 N

Initial speed = 1.5 m/s

Final speed = 3.0 m/s

We need to calculate the acceleration

Using balance equation

$mg\sin\theta-F=ma$

Put the value into the formula

$30\times10\sin15-50=30\times a$

$a=\dfrac{30\times10\sin15-50}{30}$

$a=0.92\ m/s^2$

We need to calculate the length of the ramp

Using equation of motion

$v^2=u^2+2as$

Put the value into the formula

$(3.0)^2=(1.5)^2+2\times0.92\times s$

$s=\dfrac{(3.0)^2-(1.5)^2}{2\times0.92}$

$s=3.66\ m$

Hence, The length of the ramp is 3.66 m

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Topic : kinematics

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