Physics, 06.12.2020 21:20 jakiyahporter0817

Consider the heaviest box of 68 kg that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.50, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 50.0 ∘ above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.50.

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Answer from: zitterkoph

Complete Question

Consider the heaviest box of 68 kg that you can push at constant speed across a level floor, where the coefficient of kinetic friction is 0.50, and estimate the maximum horizontal force that you can apply to the box. A box sits on a ramp that is inclined at an angle of 50.0 ∘ above the horizontal. The coefficient of kinetic friction between the box and the ramp is 0.50.If you apply the same magnitude force, now parallel to the ramp, that you applied to the box on the floor, what is the heaviest box (in pounds) that you can push up the ramp at constant speed? (In both cases assume you can give enough extra push to get the

The force on the heaviest mass is $F = 333.2 \ N$

The mass on the inclined ramp is $M = 76.47 \ kg$

Explanation:

From the question we are told that

The mass of the heaviest box is $m = 68 \ kg$

The coefficient of kinetic friction is $\mu_k = 0.50$

The angle at which the ramp is inclined is $\theta = 50^o$

Generally at constant speed the acceleration is zero , which implies that

F - F_f = 0

Here F is the horizontal force applied to the heaviest box and F_f is the kinetic frictional force which is mathematically represented as

$F_f = \mu_k * m * g$

=> F_f = 0.50 * 68 * 9.8

=> $F_f = 333.2 \ N$

F - 333.2 = 0

=> $F = 333.2 \ N$

Considering the ramp inclined at an angle above the horizontal

Generally apart from the force F acting on the box in a direction parallel to the ramp the other force acting on the box parallel to the ramp is the net force due to frictional force on the object which is mathematically represented as

$F_f_h = \mu Mg cos (\theta )$

and the weight of the object which is mathematically represented as

$F_w = Mg sin (\theta )$

So the net force is mathematically evaluated as

$F_n = F_w - F_{fh}$

=> $F_n =Mg sin (\theta ) - \mu Mg cos (\theta )$

=> F_n =M [ 9.8 sin ( 50 )-0.50 * 9.8 cos (50 ) ]

=> $F_n = 4.3576M \ N$

So the net force acting on the box parallel to the ramp is mathematically represented as

$F_{net} = F - F_n$

=> $F_{net} = 333.2- 4.3575M$

=> ma = 333.2- 4.3575M

From the question we are told the velocity is constant so the acceleration is zero

=> 333.2- 4.3575M = 0

=> $M = 76.47 \ kg$

Answer from: Quest

the answer is to check all of them! they all apply well at least i think so!

Answer from: Quest

0.55x745.7j and a calculator

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Consider the heaviest box of 68 kg that you can push at constant speed across a level floor, where t...