9. its speed at the positions x-100m is (1 pt) a. O m/s b. 2.24 m/s c. 3.87 m/s

Question

9. its speed at the positions x-100m is (1 pt) a. O m/s b. 2.24 m/s c. 3.87 m/s d. 3.16 m/s H) A spring is hung vertically, and an object of mass m is attached to its lower end. Under the action of the "load mg, the spring stretches a distance d from its equilibrium position. 10. If a spring is stretched 1.00 cm by a suspended object having a mass of 1.00 kg, the force constant of the spring is(1.0 pt) a. 1.00x10 N/m b. 9.80 N/m c. 98.0 N/m d. 9.80x102 N/m 11. Work done by the spring on the object as it stretches through this distance is(1.0 pt) a. 0.0490 b. 0.490 J c. 4.90 J d. 490 J l) A 4.0-kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of 12 N. 12. Find the speed of the block after it has moved 1.9 m if the surfaces in contact have a coefficient of kinetic friction of 0.14 (1.0 pt) Ax ng a. 1.5 m/s b. 3.5 m/s c. 4.8 m/s d. 2.5 m/s Ax ing

Explanation / Answer

Q10.

Given:

mass of the suspended object = m = 1 kg

Weight of the suspended object =mg = 1 x 9.8 = 9.8 N

stretching of the spring = x = 1.0 cm =0.01 m

force constant of the spring = mg / stretching of the spring = 9.8 /0.01 = 9.8 x 102 N/m

answer is option D

Q11. Work done by the spring as it stretches through the distance = 1/2 x spring constant x stretch 2

= 1/2 x 9.8 x 102 x (0.01)2

= 0.049 J

answer is Option a

Q12.

Horizontal force acting on the block = 12 N

frictional force = 0.14 x 4 x 9.8 = 5.488 N

Net force acting on the block = 12 - 5.488 = 6.512 N

acceleration of the block = Net force / mass =6.512/4= 1.628 m /s2

initial velocity = vi = 0

final velocity = vf = ?

acceleration = a = 1.628 m /s2

distance covered= d= 1.9 m

applying equation of motion

vf2 - vi2 = 2ad

so, vf2 - 0 = 2x1.628 x 1.9

so, vf = 2.487 m/s

answer is option d

all the best in the course work

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