Answer to Question #240467 in Mechanics | Relativity for Anas

Question #240467

A block slides down a frictionless plane having an inclination of 0 = 15.0°. The block starts from rest at the top, and the length of the incline is 2.00 m. (a) Draw a free-body diagram of the block. Find (b) the acceleration of the blockand (c) its speed when it reaches the bottom of the incline


1
Expert's answer
2021-09-23T15:48:56-0400

(a) The figure represents the free-body diagram of the block that slides on frictionless inclined surface.



(b) The weight of the block is resolved into two components. The component

mg cos θ

is equal to the normal force and the component

mg sin θ

is the force that slides the block on the inclined surface. Here

θ

is angle of inclination. From Newton’s law the net force is equal to the product of mass

(m)

and acceleration

(a)

Therefore,

Fnet=mgsinθma=mgsinθa=gsinθF_{net} = mg sin θ \\ ma=mg sin θ \\ a = g sin θ

The angle of inclination,

θ = 15.0 °

Therefore acceleration of the block is:

=9.80×sin15.0°=2.54  m/s2= 9.80 \times sin 15.0° \\ = 2.54 \;m/s^2

We can find the speed of the block when it reaches the bottom of the incline by using kinematic equation,

vf2vi2=2asv_f^2 -v_i^2 = 2as

v_f is the final speed of the block at the bottom of incline

s is the length of the incline

The block starts from rest at the top of inclined surface. So, the initial speed of the block at the top of inclined surface,

vi=0v_i=0

Length of the inclined surface,

s=2.00 m

Acceleration of the block,

a=2.54  m/s2a=2.54 \; m/s^2

Substituting the values in above equation we can find the speed of the block when it reaches the bottom of incline.

vf2vi2=2asvf20=2(2.54)(2.00)vf=2×2.54×2.00=3.18  m/sv_f^2 -v_i^2 = 2as \\ v_f^2 -0 = 2 (2.54)(2.00) \\ v_f = \sqrt{2 \times 2.54 \times 2.00} \\ = 3.18 \;m/s


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