The figure shows the force (F) versusdisplacement (s) graph for a part

1.	The figure shows the force (F) versusdisplacement (s) graph for a particle of mass m= 2kg initially at rest
Answer» Answer: (1) Maximum speed occurs at x=3 (2) Maximum speed is V=4.183 m/s (3) At x=6, V=0. Therefore correct option is (B). Explanation: We will use the concept of kinetic energy. Since it is the only FORMULA which relates FORCE, distance, and area under the curve. The kinetic energy at a GIVEN moment is equal to the work done from the beginning to that moment. K.E= (1/2)x(M)x(V^2) Area under the curve upto X=3 is, 17.5 Joule (J). Mathematically, Area= 2.5+2.5+5+5+2.5 = 17.5 J Therefore, K.E at x=3 is 17.5 J using formula, 17.5 J= (1/2)x(2)×(V^2) V= 4.183 m/s (For your practise, you can try to find velocity at different VALUES of X.) ALSO at X=6, K.E=0 J Therefore, using formula 0 =(1/2)x(2)x(v^2) V= 0 m/s Also, as additional bonus let's find total kinetic energy. Area under the curve= - 2.5 J

The figure shows the force (F) versusdisplacement (s) graph for a particle of mass m= 2kg initially at rest

Answer»

Answer:

(1) Maximum speed occurs at x=3

(2) Maximum speed is V=4.183 m/s

(3) At x=6, V=0.

Therefore correct option is (B).

Explanation:

We will use the concept of kinetic energy. Since it is the only FORMULA which relates FORCE, distance, and area under the curve.

The kinetic energy at a GIVEN moment is equal to the work done from the beginning to that moment.

K.E= (1/2)x(M)x(V^2)

Area under the curve upto X=3 is, 17.5 Joule (J).

Mathematically,

Area= 2.5+2.5+5+5+2.5 = 17.5 J

Therefore, K.E at x=3 is 17.5 J

using formula,

17.5 J= (1/2)x(2)×(V^2)

V= 4.183 m/s

(For your practise, you can try to find velocity at different VALUES of X.)

ALSO at X=6,

K.E=0 J

Therefore, using formula

0 =(1/2)x(2)x(v^2)

V= 0 m/s

Also, as additional bonus let's find total kinetic energy.

Area under the curve= - 2.5 J

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