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Tanzania [10]
3 years ago
14

A box slides down a 30.0° ramp with an acceleration of 1.20 m/s^2. Determine the coefficient of kinetic friction between the box

and the ramp.

Physics
1 answer:
Zielflug [23.3K]3 years ago
4 0

m = mass of the box

N = normal force on the box

f = kinetic frictional force on the box

a = acceleration of the box

μ = coefficient of kinetic friction

perpendicular to incline , force equation is given as

N = mg Cos30                                         eq-1

kinetic frictional force is given as

f = μ N

using eq-1

f = μ mg Cos30    


parallel to incline , force equation is given as

mg Sin30 - f = ma

mg Sin30 - μ mg Cos30  = ma

"m" cancel out

a = g Sin30 - μ g Cos30

inserting the values

1.20 = (9.8) Sin30 - (9.8) Cos30 μ

μ = 0.44

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The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm and a ball with
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B) If there is friction, the total work is;

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Work of the ball is;

Wb = KE + Wf

So, Wb = ½mv² + fx - - - (2)

Combining both equations, we have;

½mv² + fx = ½kx²

Plugging in the relevant values, we have;

(½ × 0.03 × v²) + (6 × 0.06) = ½ × 400 × 0.06²

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So, from F = Kx;

(x is measured into barrel from end where F = 0)

Thus; 6.0 = 400x

x_m = 6/400

x_m = 0.015m from the end after traveling 0.045m

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[(400 x 0.06) - 6.0] = 18N

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(v_max)² = 2 x 0.405/0.03

(v_max)² = 27

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