of 2.00 m/s2 in the positive x- direction, Neglecting frictional forces on the t

Question

of 2.00 m/s2 in the positive x- direction, Neglecting frictional forces on the traller, determine the folowing. (Indicabe the direction with the sign of Take the forward direction to be pesitive. Assume the trailer's weight is entirely supported by it's own tires ) (a) the net force on the car (b) the net force on the traler (c) the force exerted by the traler on the car The response you submitted has the wrong sign. N (d) the resultant force exerted by the car on the read magntude2130 jv'N direction 75. (below the Need Help? A1.460-N cate is being pushed horizontal, as shown in the figure oss a level floor at a constant speed by a free ronso s at an angle of 20.0s bean the (a) below 20 (a) What is the coefficient of kinetic friction between the crate and the floor? (b) It the 290-N force is instead puiling the block at an angle of 20.0 above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coeficient of friction is the same as that found in part (a) Your response differs from the correct answer by more than 10%. Double check your

Explanation / Answer

a) Net force on the car is Fcar = m*a = 1200*2 = 2400 N

b) The net force on the trailer = m*a = 305*2 = 610 N

c) force exerted by the trailer on the car = -305*2 = -610 N

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4) a) frictional force = F*cos(20)

mu_k[(F*sin(20)+mg)] = F*cos(20)

mu_k = F*cos(20)/[(F*sin(20)+mg)]

mu_k = (290*cos(20))/(290*sin(20)+1460)

mu_k = 0.174

b) frictional force - F*cos(20) = Fnet

mu_k*[mg - F*sin(20)] - F*cos(20) = m*a

0.174*(1460-(290*sin(20))) - 290*cos(20) = (1460/9.8)*a

accelaration is a = 0.24 m/s^2

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