A block of mass 0.5 kg is pushed against a horizontal spring of negligible mass,

Question

A block of mass 0.5 kg is pushed against a horizontal spring of negligible mass, compressing the spring a distance of x as shown in Figure 5. The spring constant is 450 N/m. When released the block travels along a frictionless horizontal surface to point B, the bottom of a vertical circular track of radius R = 1 m and continues to move up the track. The circular track is not smooth. The speed of the block at the bottom of the vertical circular track is vB = 12 m/s, and the block experiences an average frictional force of 7.0 N while sliding up the curved track.

What was the initial compression of the spring?

What is the speed of the block at the top of the circular track?

Does the block reach the top of the track, or does it fall off before reaching the top?

Explanation / Answer

A) Apply conservation of energy

initial elastic potentail energy = kientic energy of block at th bottom of the circular track.

0.5*k*x^2 = 0.5*m*vB^2

x^2 = m*vB^2/k

x = vB*sqrt(m/k)

= 12*sqrt(0.5/450)

= 0.4 m <<<<<<<<----------Answer

B) Workdone by gravity on the block as it slides to the top of the loop, W_gravity = -m*g*(2*R)

= -0.5*9.8*2*1

= -9.8 J

Workdone by friction, W_friction = -fk*d

= -fk*pi*R

= -7*pi*1

= -22 J

net workdne, Wnet = W_gravity + W_friction

= -9.8 -22

= -31.8 J
Kinetic energy of the block at thr bottom, KEi = 0.5*m*Vb^2

= 0.5*0.5*12^2

= 36 J

KEi > |workdone|. so the block will reach top of the loop. <<<<<<<<----------Answer

let v is speed of block at the top of the loop

Now Apply work-energy theorem.

Wnet = KEf - KEi

-31.8 = 0.5*m*vf^2 - 36

36 - 31.8 = 0.5*0.5*v^2

4.2 = 0.25*v^2

==> v^2 = 4.2/0.25

v = sqrt(16.8)

= 4.1 m/s <<<<<<<<----------Answer

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