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usn39or.theexpertta.com + Google Current Studen Portland Conco Portland Conco- Not Signed In l..The Expert Shudent: dscstaltingsigmall.com My Account Log Out h. S Homemerk Begin Date: 00 PM End Date: 12/14/2017 12:00:00 AM i winter spoeting event pulls a 4kg block of ice n the positive horizontal direction with a rope over his shoulders across a frenen lake as shown in the figure. Assunle the coefficients of statsic and kinetic friction ane ,-0.1 and p0.03 25 Ctheespertta 50% Part (a) Calculate the milit um fece F he must exert to get the block sliding innewwa Grade Semmary Deluctions sino cos) tan) sinho ODegrees Radians Hiats:t deduction per hint. ies remaining 50% Part (b) What is its acceleration i mi, oce it starts to move, if that force smantaned?Explanation / Answer
Part a
The normal force of the lake ice on the block is
Fn = F sin(theta) + mg
The friction force before sliding is
Ff = u Fn
Ff = u(Fsin(theta) + mg)
So,
the force needed to begin to displace the block is(using the static friction coefficient) is equal to the friction force
But we also know from geometry that
Fh = Fcos(theta)
Fcos(theta) = us(Fsin(theta) + mg)
Fcos(theta) = us Fsin(theta) + us mg
Fcos(theta) - us Fsin(theta) = us mg
F(cos theta - ussin theta) = usmg
F = usmg/(cos theta - ussin theta)
F = smg / (cos theta - ssintheta)
= 0.1*44*9.8/(cos 25 degree - 0.1* sin 25 degree)
= 50 N
Part b
Now, with a force of 50 N what is the acceleration with coefficient of friction at 0.03
So, the net force Fa accelerating the block is the total horizontal force minus the friction force
Fa = Fh - Ff
Fa = Fcostheta - uk(Fsin theta + mg)
As Fa = ma
ma = Fcos theta - uk(Fsin theta + mg)
a = Fcos theta - uk(Fsin theta + mg) / m
= (50*cos 25 degree - 0.03*(50*sin 25 degree + 44*9.8))/44
= 0.72 m/s^2
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