A 2.5-kg mass starts from rest at point A and moves along the x axis subject to

Question

A 2.5-kg mass starts from rest at point A and moves along the x axis subject to the potential energy shown in the figure below.

PE 30 J 25 20 J 15 J 10 J 5 J 2 m 4 m 6 m 8 m 10 m (a) Determine the speed of the mass at points B, C, D. point B m/s point C m/s 82 1.86 x point D Is energy conserved? If so, how do you know it is conserved? How does the total energy at A compare to the total energy at any other point. m/s (b) Determine the turning points for the mass. (Select all that apply.) point A point B point C point D point E

Explanation / Answer

In this the solution of given values....

The total Mechanical Energy at point A is E = PE + KE

T M E = 30 J

As it starts from rest, Ke = 0.

Thus the value of E are .....

E = PE = 30 J

by conservation of mechanical energy.the total mechanical energy is 30 J,

Thus the point of value ...,

At point B:

if PE = 5 J

then

KE = 25 J = 0.5 mv^2

so V^2 = 2KE/mb

V^2 = 2*25/2.5

V at B = 4.47 m/s

At point C:

PE = 20 J

so

KE = 0.5 mv^2 = 10 J.

Vc^2 = 2 K/m

Vc^2 = 2 * 10/2.5

Vc = 2.82 m/s

At point D:

PE = 15 J

So value it will be.....

KE = 30 - 15 = 15 J

0.5 mv^2 = 15

V^2 = 2*15/2.5

V at D = 1.86 m/s

at Max height, V= 0 will be the turning Point

A and E here it is

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