Question

4.A child of mass 30kg skates down the top of a ramp having a constant slope of 15^(@) .The child's speed increases from 1.5ms^(-1) to 3.0ms^(-1) as she reaches the bottom of the ramp.A force of kinetic friction of magnitude 50N opposes her mottion.Draw the appropriate force diagram and determine the length of the ramp. Take g=10ms^(-2) .​

Answer 1

Answer:

The length of the ramp will be 3.67 m

Explanation:

As we know that there are two forces along the ramp on the child

1) Down the ramp child has gravitational force

2) Along the ramp upwards child has frictional force

so we will have net force on child given as

$F_g sin\theta - F_f = ma$

$a = gsin\theta - \frac{F_f}{m}$

$a = 10 sin15 - \frac{50}{30}$

$a = 2.59 - 1.67$

$a = 0.92 m/s^2$

now we have

$v_f^2 - v_i^2 = 2 a d$

$3^2 - 1.5^2 = 2(0.92) d$

$d = 3.67 m$

#Learn

Topic : Motion of inclined plane

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