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प्रश्न
In a children-park an inclined plane is constructed with an angle of incline 45° in the middle part (in the following figure). Find the acceleration of boy sliding on it if the friction coefficient between the cloth of the boy and the incline is 0.6 and g = 19 m/s2.

योग
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उत्तर
Let m be the mass of the boy.
From the above diagram:
R − mg cos 45° = 0
R = mg cos 45° =`"mg"/sqrt2` (1)
Net force acting on the boy, making him slide down
= mg sin 45° − μR
= mg sin 45° − μmg cos 45°
`=mxx10xx(1/sqrt2)-0.6xxmxx10xx(1/sqrt2)`
`=m(5sqrt2-3sqrt2)=mxx2xxsqrt2`
The acceleration of the boy = `"Force"/"Mass"`
`=(m(2sqrt2))/m`
= 2√2 m/s2
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