Question

An elephant (mass = 5 tonne) takes tourists to a hill resort 0.4 km above the ground level. The elephant Carries a 2000kg load of tourists and luggage and takes 60 minutes. Calculate the i. Work done by the elephant ii. Potential energy gained by the tourists and luggage iii. Efficiency the operation iv. Power of the elephant in horsepower​

Answer 1

Answer:

your answer is in this pix

Answer 2

i) $W=27440000\ J=2.744\times 10^7\ J$

ii) $U=7840000\ J=7.84\times 10^6\ J$

iv) $P=7622.22\ W =10.2175\ hP$

Explanation:

Given:

• mass of elephant, $m=5\ tonne=5000\ kg$

• displacement climbed by the elephant, $s=0.4\ km=400\ m$

• mass of tourist and luggage, $m_l=2000\ kg$

• time taken to climb, $t=60\ min.=3600\ s$

i)

work done by the elephant:

$W=(m+m_l)\times g\times s$

$W=(5000+2000)\times 9.8\times 400$

$W=27440000\ J=2.744\times 10^7\ J$

ii)

Potential gained by the tourist and the luggage:

$U=m_l.g.s$

$U=2000\times 9.8\times 400$

$U=7840000\ J=7.84\times 10^6\ J$

iv)

Power of the elephant:

$P=\frac{W}{t}$

$P=\frac{27440000}{3600}$

$P=7622.22\ W =10.2175\ hP$

iii)

The efficiency of an inclined plane is given as:

$\eta=\frac{AMA}{IMA}$

where:

IMA = ideal mechanical advantage

AMA = actual mechanical advantage

IMA is calculated by:

$IMA=\frac{l}{h}$

where:

$l=$ inclined length of the slope

$h=$ final height of the inclination at the end of the slope.

AMA is calculated by:

$AMA=\frac{m.g}{m.g\ sin\ \theta} =\frac{1}{sin\ \theta}$

Actual mechanical advantage is the ratio of the force applied for without the use of inclined plane to the force applied using inclined plane.

Here we are not given with the required data for calculation.

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TOPIC: inclined plane

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