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sweet [91]
4 years ago
13

A carton is given a push across a horizontal, frictionless surface. The carton has a mass m, the push gives it an initial speed

of vi, and the coefficient of kinetic friction between the carton and the surface is μk.
(a) Use energy considerations to find an expression for the distance the carton moves before it stops. (Use any variable or symbol stated above along with the following as necessary)
(b) What if the initial speed of the carton is increased by a factor of 3, determine an expression for the new distance d the box slides in terms of the old distance.
Physics
1 answer:
sammy [17]4 years ago
8 0

Answer and Explanation:

Data provided in the question

Carbon mass = m

Initial speed = v_i

Coefficient = μk

Based on the above information, the expressions are as follows

a. By using the energy considerations the expression for the carton moving distance is

As we know that

Fd = \frac{1}{2} m (v_i^2- v_f^2)

where,

v_f = 0

F = u_kmg

(\mu_kg) d = \frac{1}{2} m v_i^2

d = \frac{\frac{1}{2}v_i^2}{\mu_kg}

d = \frac{v_i^2}{2 \mu_kg}

b. The initial speed of the carton if the factor of 3 risen, so the expression is

v_i^1 = 3v_i

d^i = \frac{(3v_i^2)}{2\mu_kg}

= \frac{9v_i^2}{2\mu_kg}

d^i = 9(d)

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