Question

A 70 kg man is being pulled to safety by a rope over a dangerous pit of alligators. If the tension in the rope is 800 N, what is the acceleration and the direction on the man?

Answer 1

Answer:

1.62 m/s²

Explanation:

$\underline{\bigstar\:\textsf{Given:}}$

• Mass (m) = 70 kg
• Tension in the rope (T) = 800 N

$\underline{\bigstar\:\textsf{To\:find:}}$

• Acceleration (a) = ?

$\underline{\bigstar\:\textsf{Solution:}}$

We know, in the cases of pulling a rope the $\blue{\sf{Tension\:=\:mg\:+\:ma}}$

Here, we know the mass (m), acceleration due to gravity (g) as 9.8 m/s²and the tension (T), we've to find acceleration (a).

$\:\:\:\::\implies$ 800 = 70 × (9.8 + a)

$\:\:\:\::\implies\:\sf{\dfrac{800}{70}\:=\:9.8\:+\:a}$

$\:\:\:\::\implies\:\sf{11.42\:=\:9.8\:+\:a}$

$\:\:\:\::\implies\sf{11.42\:-\:9.8\:=\:a}$

$\:\:\:\::\implies\:\sf{1.62\:=\:a}$

$\:\:\:\:\small:\implies\underline{\boxed{\sf\purple{a\:=\:1.62\:m/s^2}}}$

Answer 2

$\huge \fbox {\fbox{\red{Answer}}}$

$\large \fbox{Given:}$

$\sf \longrightarrow{Mass \: of \: man(m) = 70kg}$

$\sf \longrightarrow{Tension \: in \: the \: rope(T) = 800N}$

$\large \fbox{To \: find:}$

$\sf \leadsto{Acceleration of \: man(a).}$

$\large \fbox{Formula used:}$

$\sf \mapsto{Tension = mg + ma}$

$\large \fbox{Concept used:}$

$\sf\longmapsto{Acceleration \: due \: to \: gravity \: is \: 9.8m/ {s}^{2}.}$

$\large \fbox{According to Question:}$

$\sf \implies{Tension = mg + ma}$

$\sf \implies{800 N= (70kg)(9.8m /{s}^{2} ) + (70kg)(a)}$

$\sf \implies{800 N= 70kg(9.8m /{s}^{2} + a)}$

$\sf \implies{ \frac{800 N}{70} = (9.8m /{s}^{2} + a)}$

$\sf \implies{11.42 N = 9.8m /{s}^{2} + a}$

$\sf \implies{ {11.42 } - {9.8m /{s}^{2}} = a}$

$\sf \implies{ {1.62m /{s}^{2}} = a}$

$\boxed{ \bf{Acceleration \: of \: man \: is \: 1.62m/{s}^{2}.}}$