A 2.0 kg block sits on a 4.0 kg block that is on a frictionless table. The coeff

Question

A 2.0 kg block sits on a 4.0 kg block that is on a frictionless table. The coefficients of friction between the blocks are µs = 0.30 and µk = 0.20.

(a) What is the maximum force F that can be applied to the 4.0 kg block if the 2.0 kg block is not to slide?
- N

(b) If F is half this value, find the acceleration of each block.
- m/s2 (2.0 kg block)
- m/s2 (4.0 kg block)
Find the magnitude of the force of friction acting on each block.
- N (2.0 kg block)
- N (4.0 kg block)

(c) If F is twice the value found in (a), find the acceleration of each block.
- m/s2 (2 kg block)
- m/s2 (4 kg block)

Explanation / Answer

(a)

m1 = 2 kg

m2 = 4 kg

for m1

frictional force f1 = us*m1*g

Fnet = m1*a

Fnet = f1

m1*a = us*m1*g ......(1)

for m2

Fnet = F - f1

but

Fnet = m2*a

m2*a = F - us*m1*g .....(2)

solving 1 & 2

2/1

m2/m1 = (F - us*m1*g)/ us*m1*g

F = us*m1*g + (m2/m1)*(us*m1*g)

F = us*m1*g + us*m2*g

F = (m1+m2)*us*g

F = (2+4)*0.3*9.8

F = 17.64 N = 18 N

=========================

(b)

If F is half

Fnet = F/2 = 9.0 N

The two masses move together

anet= Fnet/(m1+m2) = 1.5 m/s^2

================

(c)

If F is doubled

the two blocks slide o one another

for m1

f1 = uk*m1*g

a1 = f1/m1 = uk*g = 0.2*9.8 = 1.96 m/s^2 = 2.0 m/s^2

for m2

Fnet = F - fk1 = 36 - (2*0.2*9.8) = 32.08

a2 = Fnet/m2 = 8.0 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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