A spring (k = 75N/m) has an equilibrium lenght of 1.00m. The springs is compress

Question

A spring (k = 75N/m) has an equilibrium lenght of 1.00m. The springs is compressed to a lenght of .50m and a mass of 2.0kg is placed at it's free end on a frictionless slope which makes an angle of 41 degrees with respect to the horizontal. The spring is then released.(a) If the mass is not attached to the spring, how far up the slope will the mass move before coming to rest?(b)If the mass is attached to the spring, how far up the slope will the mass move before coming to rest?(c)Now the incline has a coefficienct of kinetic friction uk, if the block attached to the springs is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction uk?

Explanation / Answer

Given that spring constant k = 75 N /m Length L = 1.00m compressed length x = 0.5 m mass m = 2.0 kg a ) 1/2 kx^2 = 1/2 mv_0^2 75 * ( 0.5 m)^2 = 2.0 kg * v_0 ^2 v_0 = 3.06 m/s using kinematic equation a = - g sin 41 = - 6.42 m/s^2 v^2 - v_0 ^2 = 2 a s 0 - ( 3.06 m/s )^2 = 2 * (-6.42 m/s^2 ) * s s = 0.728 m b ) from law of conservation of energy distance is d = 0.5 m c ) coefficient of frictionb is µ mg cos 41 * x = 1/2 kx^2 µ * 2.0 * 9.8 *cos (41) 0.5 = 1/2 *75 * (0.5 )^2 µ = 1.267 PLEASE RATE!!

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