2. A monkey of mass 40kg climbs on a rope which can stand a maximum te

1.	2. A monkey of mass 40kg climbs on a rope which can stand a maximum tension on 600N. Inwhich of the following cases will the rope break: the monkey(a) Climbs up with an acceleration of 6 ms2(b) Climbs down with an acceleration of 4 ms2(c) Climbs up with a uniform speedf ms(d) Falls down the rope nearly freely undergravity
Answer» Answer:- Case (a) Mass of the monkey,m= 40 kgAcceleration due to gravity,g= 10 m/sMaximum tension that the rope can bear,Tmax= 600 NAcceleration of the monkey,a= 6 m/s² upwardUsing Newton’s second law of motion, we can write the equation of motion as:T–mg= ma∴T=m(g+ a)= 40 (10 + 6)= 640 NSinceT>Tmax, the rope will break in this case. Case (b)Acceleration of the monkey,a= 4 m/s2downwardUsing Newton’s second law of motion, we can write the equation of motion as:mg –T=ma∴T=m(g– a)= 40(10 – 4)= 240 NSinceT<Tmax, the rope will not break in this case. Case (c)The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e.,a= 0.Using Newton’s second law of motion, we can write the equation of motion as:T–mg= maT–mg = 0∴T=mg= 40 × 10= 400 NSinceT<Tmax, the rope will not break in this case. Case (d)When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e.,a =gUsing Newton’s second law of motion, we can write the equation of motion as:mg –T=mg∴T=m(g–g) = 0SinceT<Tmax, the rope will not break in this case.

2. A monkey of mass 40kg climbs on a rope which can stand a maximum tension on 600N. Inwhich of the following cases will the rope break: the monkey(a) Climbs up with an acceleration of 6 ms2(b) Climbs down with an acceleration of 4 ms2(c) Climbs up with a uniform speedf ms(d) Falls down the rope nearly freely undergravity

Answer»

Answer:-

Case (a)

Mass of the monkey,m= 40 kgAcceleration due to gravity,g= 10 m/sMaximum tension that the rope can bear,Tmax= 600 NAcceleration of the monkey,a= 6 m/s² upwardUsing Newton’s second law of motion, we can write the equation of motion as:T–mg= ma∴T=m(g+ a)= 40 (10 + 6)= 640 NSinceT>Tmax, the rope will break in this case.

Case (b)Acceleration of the monkey,a= 4 m/s2downwardUsing Newton’s second law of motion, we can write the equation of motion as:mg –T=ma∴T=m(g– a)= 40(10 – 4)= 240 NSinceT<Tmax, the rope will not break in this case.

Case (c)The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e.,a= 0.Using Newton’s second law of motion, we can write the equation of motion as:T–mg= maT–mg = 0∴T=mg= 40 × 10= 400 NSinceT<Tmax, the rope will not break in this case.

Case (d)When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e.,a =gUsing Newton’s second law of motion, we can write the equation of motion as:mg –T=mg∴T=m(g–g) = 0SinceT<Tmax, the rope will not break in this case.

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