(a) A crate of mass M. 61.9kg rests on a level surface, with a coefficient of ki

Question

(a) A crate of mass M. 61.9kg rests on a level surface, with a coefficient of kinetic friction N .98You push on the crate with an applied force of 0.840N. What is the magnitude of the crate's acceleration as it slides? Answer: m/s 670 but now place the crate on an inclined surface, (b) Take the same crate of mass M. -k& and the same coefficient of 4- kinetic friction N 008 slanted at some angle above the horizontal. Now instead of pushing on the crate, you let it slide down due to gravity. What must the angle 0 be, in order for the crate to slide with the same acceleration it had in part (a)? Answer: degrees kinetic friction N: angle be, in order for the crate to slide with the same acceleration it had in part (a)? Answer: (c) Same situation as in part (b), but now with no friction. Once again, what must the ._degrees

Explanation / Answer

here,

Part A:
mass of crate, mc = 67.9 kg
coefficient of friction, u = 0.98

From Newton Second Law, Fnet = ma
F - Ff = ma ( F is applied force, Ff is frictional force)
F - u*mg = ma
0.840 - 0.98*67.9*9.81 = 67.9 * a ( a is acceleration)

Therefore acceleration, a = -9.601 m/s ( - shows deacceleration )

Part B:
From Newton Second Law, SUM(F) = 0
mgsinA - Ff = 0 ( acceleration, a = 0 since there is no applied force)
mgSinA = Ff ( Ff is frictional force)
mgSinA = u*mg*CosA
TanA = u
Angle, A = arcTan(u) = arcTan(0.98) = 44.421 degrees

Part C:
From Newton Second Law, Fnet = ma
mgSinA = ma
Angle, A = arcSin(a/g)
A = arcSin(9.601/9.81)
A = 78.152 degrees

Part D:
Radius, r = 326 m

Since Acceleration in circular motion,
a = g = v^2/r

solving for velocity, v
v = sqrt(g*r)
v = sqrt(9.8*326)
v = 56.523 m/s

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