Probleml Two bricks with masses mi-5 kg and m-3 kg lie at rest on a frictionless

Question

Probleml Two bricks with masses mi-5 kg and m-3 kg lie at rest on a frictionless horizontal plane. The bricks are attached to each other by an ideal string of length d-0.035 m. An ideal mass-less spring with spring constant k-2500 N/m and relaxed length D-0.05 m is compressed and then inserted between the bricks as shown in Fig. 1 and as we did in class. (a) Calculate the tension in the string. (b) The string is now cut and consequently the bricks are pushed apart by the spring. Calculate the final velocity of each brick following this event. 2500 N/m m1= 5 kg m2 3 kg Figure 1

Explanation / Answer

a)

Applying Newton’s 2nd law we get,

T=F

T=k*x

Plugging values,

T=2500*(0.05-0.035)

T = 37.5 N

b)

By law of conservation of momentum,

m2v2-m1v1 = 0

m1v1=m2v2

v2=(m1/m2*v1 -------------(1)

By law of conservation of energy,

PEi=KEf

1/2k*x^2 = 1/2m1v1^2+1/2m2v2^2

½*2500*(0.050-0.035)^2=1/2*5.0*v1^2+1/2*3.0*[(5.0/3.0)*v1]^2

v1=0.205 m/s

Plug in (!),

v2=(5.0/3.0)*0.205

v2=0.34 m/s

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.