Question

A plank with a box on it at one end is gradually raised about the other end. As the angle of inclination with the horizontal reaches 30° the box starts to slip and slides 4.0 m down the plank in 4.0s. The coefficients of static and kinetic friction between the box and the plank will be, respectively:(a) 0.6 and 0.5(b) 0.5 and 0.6(c) 0.4 and 0.3(d) 0.6 and 0.6

Answer 1

Answer:

a) 0.6 and 0.5

Explanation:

Let μk and μs be the coefficients of static and kinetic friction between the box and the plank.  

When the angle of inclination 0 reaches 30°, the block will slide  

Therefore μs = tanθ = tan30° = 1/√3 =0.6  

If a is acceleration produced in the block, then  

ma=mgsinθ;−fk  (where fk = force of kinetic friction)  

= mgsinθ;−μhN (as fk = μkN)  

= mgsinθ;−μhmgcosθ (as N=mgcosθ )  

a = g(sinθ−μhcosθ)  

Asg = 10ms² and θ = 30°  

Therefore, a = (10ms²)(sin30∘−μhcos30∘) -- 1

If s is the distance traveled by the block in time t, then  

s =1/2at² or a=2st²  

But s - 4.0m and t = 4.0s ( Given)  

Thus a = 2(4.0m)/(4.0s)² = 1/2ms−2  

Substituting the value of a in eq 1

1/2ms−2 = (10ms−2)(1/2−μk√3/2)  

1/10 = 1−√3μk

or √3μk = 1−1/10 = 9/10 = 0.9  

μk = 0.9/√3 =0.5

Thus, the coefficients of static and kinetic friction between the box and the plank will be, respectively 0.6 and 0.5

Answer 2

Answer:

0.6&0.5

Explanation:

chk the pic for detailed explanation