Conservation of Energy

**MarcoMP** · November 13th 2009, 01:48 AM

There is a 3Kg crate at the bottom of an incline, initially moving at $3 m.s^{-1}$ . The incline is 20 degrees and frictionless. The crate moves through a displacement of d metres and comes to a stop. Calculate d.

I did the following:

$(E_{p} + E_{k})_{Bottom} = (E_{p} + E_{k})_{Top}$
$0 + {1/2}.(3).(3)^2 = (3).(9.8).Sin20.d + 0$
$d = 1.34m$

Is this correct?

**mathaddict** · November 13th 2009, 03:02 AM

Originally Posted by MarcoMP

There is a 3Kg crate at the bottom of an incline, initially moving at $3 m.s^{-1}$ . The incline is 20 degrees and frictionless. The crate moves through a displacement of d metres and comes to a stop. Calculate d.

I did the following:

$(E_{p} + E_{k})_{Bottom} = (E_{p} + E_{k})_{Top}$
$0 + {1/2}.(3).(3)^2 = (3).(9.8).Sin20.d + 0$
$d = 1.34m$

Is this correct?

HI

From the principle of conservation of energy , PE gained = KE lost

$\frac{1}{2}mu^2=\frac{1}{2}mv^2+mgh$

$\frac{1}{2}(3)(3)^2=0+(3)(9.81)(d\sin 20)$

$d=1.341$

I agree with your answer .

**Raoh** · November 13th 2009, 03:02 AM

looks fine to me.

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