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April 17, 2014

April 17, 2014

Posted by **phys1!** on Friday, October 14, 2011 at 10:46pm.

of 18 kg and is pulled at constant speed by

a rope. The rope is inclined at 30.4 above

the horizontal and the cart moves 21.5 m on

a horizontal floor. The coefficient of kinetic

friction between ground and cart is 0.8 .

The acceleration of gravity is 9.8 m/s2 .

How much work is done on the cart by the

rope?

Answer in units of kJ

- physics -
**drwls**, Saturday, October 15, 2011 at 1:55amCalculate the rope tension from the information given.

Call it T.

The horizontal component of T = Friction force

since it does not accelerate.

(M*g - T sin21.5) *0.8 = T cos 21.5

Solve for T

Work done by rope = T*cos30.4*21.5 m

- physics -
**Kaspar**, Tuesday, February 28, 2012 at 10:48pmT*cos(θ) = Ffr –––> Ffr = µk* Fn –––>

Fn = m*g - T*sin(θ) :

T*cos(θ) = ( µk*(m*g - T*sin(θ) )

T*cos(30.4º) = (0.8 *(18*9.8 – T*sin(30.4º))

T*cos(30.4º) = 88.2 N – 0.8 * T *sin(30.4º)

T*cos(30.4º) + 0.8*T*sin(30.4º) = 141.12 N

T (cos(30.4º) + 0.8*sin(30.4º)) = 141.12 N

T = 141.12 N / (cos(30.4º) + 0.8*sin(30.4º))

T = 111.4 N

-------------------------

Work = T*cos(θ)*d

W = 111.4*cos(30.4º)*21.5m = 2065.81 J or 2.07 kJ

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