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Blizzard [7]
3 years ago
8

A 4.0-kg mass is moving with speed 2.0 m/s. A 1.0-kg mass is moving with speed 4.0 m/s. Both objects encounter the same constant

braking force, and are brought to rest. Which object travels the greater distance before stopping?
Physics
1 answer:
Kazeer [188]3 years ago
6 0

Answer

given,

mass of object 1 = 4 Kg

Speed of object 1 = 2 m/s

mass of object 2 = 1 Kg

speed of object 2 = 4 m/s

KE of the object 1 = \dfrac{1}{2}MV^2

                             = \dfrac{1}{2}\times 4 \times 2^2

                             = 8 J

KE of the object 2 = \dfrac{1}{2}MV^2

                             = \dfrac{1}{2}\times 1 \times 4^2

                             = 8 J

Kinetic energy of both the object is same hence,Work done by both the object will also be same.

It is given that braking force is same in both cases.

So, distance travel by Both the object will be same.

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Answer:

94.67 N

Explanation:

Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.

At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.  

\begin{array}{l}\\\Sigma {F_y} = 0\\\\N - W - F\sin \theta = 0\\\\N = W + F\sin \theta \\\end{array}

Substituting 70 N for W, 41 N for F and \theta for 37 degrees

N=70+41sin37=94.67441595  N and rounding off to 2 decimal places

N=94.67 N

​

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A 25.0 kg pickle is accelerated from rest through a distance of 6.0m in 4.0s across a level floor . If the friction force betwee
SIZIF [17.4K]
Add the KE increase and the work done against friction.

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