Question

# Two blocks of mass m1=4 kg and m2=2 kg, connected by a weightless rod on a plane having inclination of 37∘ is shown in figure. If the coefficient of dynamic friction is μ=0.25, then the common acceleration of the two blocks and the tension (T) in the rod are

A
4 m/s2, T=0 N
B
2 m/s2, T=5 N
C
10 m/s2, T=10 N
D
15 m/s2, T=9 N

Solution

## The correct option is B 4 m/s2, T=0 NBoth the blocks are connected with the rod so both the blocks will moves with common acceleration. Simultaneously considering blocks and rod as a system. F.B.D. of system, From the F.B.D, N2=m2gcosθ  and  N1=m1gcosθ f2=μN2  and  f1=μN1 ∴f2=μm2gcosθ            and f1=μm1gcosθ Total frictional force on the system, f=f1+f2=μgcosθ(m1+m2) Assuming a is the common acceleration along the surface. (m1+m2)gsinθ−(m1+m2)μgcosθ=(m1+m2)a ⇒10×sin37∘−0.25×10×cos37∘=a ⇒10×35−0.25×10×45=a ⇒a=4 m/s2 Now for tension take m2 block as a system F.B.D. m2 Block f=μm2gcosθ Therefore,  m2gsinθ−f−T=m2a ⇒m2gsinθ−μm2gcosθ−T=m2a ⇒2×10×35−0.25×2×10×45−T=2×4 ⇒12−4−T=8 T=0 NPhysics

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