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yanalaym [24]
3 years ago
11

A 51.1-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.622 and 0.368,

respectively. What horizontal pushing force is required to (a) just start the crate moving and (b) slide the crate across the dock at a constant speed?
Physics
1 answer:
Charra [1.4K]3 years ago
4 0

Answer:

(a)  The force must be greater than 311.4851 N to just start the crate moving

(b)  The 184.287 N force needed to to slide the crate at constant speed      

Explanation:

We have given that mass of the crate m = 51.1 kg

Coefficient of static friction \mu _s=0.622 and coefficient of kinetic friction \mu _k=0.368

Acceleration due to gravity g=9.8m/sec^2

(A) Static frictional force is given by F_S=\mu _Smg=0.622\times 9.8\times 51.1=311.48516N

So the force must be greater than 311.4851 N to just start the crate moving

(b) For sliding the crate across the dock at a constant speed force must be equal to kinetic friction force

Kinetic friction force is given by F_k=\mu _kmg=0.368\times 9.8\times 51.1=184.287N

So the 184.287 N force needed to to slide the crate at constant speed

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R=u_xt=(ucos \theta)t

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In the time t, the net displacement of the shotput is y in the downward direction.

Use the equation of motion,

y=u_yt-\frac{1}{2}gt^2=(usin\theta) t-\frac{1}{2}gt^2

Substitute the value of t from equation (1).

y=(ucos\theta)(\frac{R}{ucos\theta} )-\frac{1}{2} g(\frac{R}{ucos\theta} )^2\\ =Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )

Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.

y=Rtan\theta-(\frac{gR^2}{2u^2cos^2\theta} )\\ (-2.10m)=(24.77 m)(tan38.0^o)-\frac{(9.8 m/s^2)(24.77m)^2}{2u^2(cos38.0^o)^2} \\ u^2=225.71(m/s)^2\\ u=15.02m/s

The shot put was thrown with a speed 15.02 m/s.




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