Please solve this physics problems

Question

A mass m = 12 kg rests on a frictionless table and accelerated by a spring with spring constant k = 4591 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μ_k = 0.48. The mass leaves the spring at a speed v = 3.8 m/s.

1)

How much work is done by the spring as it accelerates the mass?

2)

How far was the spring stretched from its unstreched length?

3)

The mass is measured to leave the rough spot with a final speed v_f = 2.2 m/s.How much work is done by friction as the mass crosses the rough spot?

4)

What is the length of the rough spot?

5)

In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length?

6)

In this new scenario, what would the coefficient of friction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch?

 

Answer 1

1) How much work is done by the spring as it accelerates the mass?

(kx²)/2 this is a potential energy of the spring which is converting to kinetic energy of the m object: so we can use energy conservation low and write:

W=(kx²)/2=(mV²)/2 form this equation we can find W=(12kg*3.82²m²)/2s²=86.64J

W=86.64J

2) How far was the spring stretched from its unstreched length?

from the equation below (as we mentioned in the answer one explanation) we can easily find the length of the stretched spring from its unstretched length that is x:

W=(kx²)/2=(mV²)/2

x=sqrt((2W)/k)= sqrt((2*86.64)/4591)=0.1942m=19.42cm Note that always you can check your answer using units. Note also the answer is approximate.

x=19.42cm.

3) The mass is measured to leave the rough spot with a final speed v_f = 2.2 m/s.How much work is done by friction as the mass crosses the rough spot?

W_f= Ekfinal-Ekinitial

Ekinitial=(mV2)/2

Ekfinal=(mVf2)/2

W_f= -57.6J

4) What is the length of the rough spot?

W_f=F_f*d*cosa= - μ_k*mg*d, note that "a" is an angle between F_f and displacement and it is 180 degree and cos180= -1

also W_f=-57.6J

from these two equations we can find d=W_f/-μ_k*d= -57.6/0.48*12*9.8=1.02m

5) In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length?

we can use find the initial speed (V₀) of the mass with the new spring X using energy conservation low:

(kx²)/2=(mV₀²)/2

Also as explained in answer 3 above:

W_f= E_kfinal-E_kinitial

E_kinitial=(mV₀²)/2

E_kfinal=(mV_f²)/2 V_f=0 since the block stops in a half-way of rough surface

W_f= -(mV₀²)/2= -(kx²)/2

W_f= F_f*(d/2)*cosa= -μ_k*mg*(d/2)

-(kx²)/2= -μ_k*mg*(d/2) from this equation we can find: x=sqrt((μ_k*mg*d)/k=0.1119m=11.19cm

Note that the answer make sense since the block travel less distance compared to question 2 when the spring was compressed more (x=19.42cm, please see question 2).

x=11.19cm.

6) In this new scenario, what would the coefficient of friction of the rough patch need to be changed to in order for the block to just barely make it through the rough patch?

new scenario means that x=11.19cm. So we have to find μk such that the block just barely make it through the rough patch. Note that just barely make it through the rough patch means that the block will stop in the end of the rough patch which means that the V_f=0.

in this case all potential energy of the spring will spend to the work done on friction force:

W_f = -(kx²)/2

W_f = - F_f*d = -μ_k*mg*d

from these two equation we can find μ_k = (kx²)/2mgd= 0.2396

μ_k = 0.2396.