Question

A force of 200N pulls a box of 10kg to the right. The coefficient of kinetic friction is 0.1 What is the horizontal acceleration of the box?
If a vertical force of 50N acts on the box vertically downwards, then what will be the acceleration of the box?(take g=10m/s2).

plz answer

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Horizontal\:acceleration=19\:m/s^{2}}}}$

$\green{\tt{\therefore{Horizontal\:acceleration=13.5\:m/s^{2}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Force(F_{1})= 200 \: N \\ \\ \tt: \implies Force(F_{2})=50 \: N \\\\ \tt: \implies Coefficient \: of \: kinetic \: friction( \mu) = 0.1 \\ \\ \tt: \implies Mass \: of \: box(m) = 10 \: kg \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Horizontal \: acceleration( a_{1}) =? \\ \\ \tt: \implies Horizontal \: acceleration( a_{2} ) =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies F_{1} - \mu N = ma \\\\ \tt\circ\:N=mg=100\:N \\\\ \tt: \implies 200 - 0.1 \times 10 \times 10 = 10 \times a \\ \\ \tt: \implies 200 - 10 = 10 \times a \\ \\ \tt: \implies 190 = 10 \times a \\ \\ \tt: \implies a = \frac{190}{10} \\ \\ \green{\tt: \implies a = 19 \: {m/s}^{2} } \\ \\ \green{\tt \therefore Horizontal \: acceleration \: of \: box \: is \: 19 \: {m/s}^{2} } \\ \\ \bold{For \: Vertical \: Force:} \\ \\ \tt \circ \: F_{1} = 200 \: N \\ \\ \tt \circ \: F_{2} = 50 \: N \\ \\ \tt: \implies F_{1} - F_{2} - \mu N= ma \\ \\ \tt\circ\:N=50+mg=150\:N\\ \\ \tt: \implies 200 - 50 - 0.1 \times 150 = 10 \times a \\ \\ \tt: \implies 135 = 10 \times a \\ \\ \tt: \implies a = \frac{135}{10} \\ \\ \green{\tt: \implies a = 13.5\: {m/s}^{2} } \\ \\ \green{\tt \therefore Acceleration \: of \: box \: is \: 13.5{ \: m/s}^{2} }$

Answer 2

$\red{\underline{\underline{\bold{Answer}}}}$

$\green{\tt{\therefore{Acceleration=19\:m/s^{2}}}}$

$\green{\tt{\therefore{Acceleration=13.5\:m/s^{2}}}}$

$\blue{\bold{Given}}$

Force = 200 N

Vertical Force = 50 N

Coefficient friction = 0.1

Mass of box = 40 KG

$\orange{\bold{To\:Find}}$

Acceleration = M?

• ATO

$\bold{\purple{By\:Newton's\:law}} \\ \tt{\orange{\implies F_{1} - \mu N = ma} }\\\\ \tt {\green{\implies 200 - 0.1 \times 10 \times 10 = 10 \times a} }\\ \\ \tt {\pink {\implies 200 - 10 = 10 \times a}} \\ \\ \tt{\orange{\implies 190 = 10 \times a}} \\ \\ \tt {\red{\implies a = \frac{190}{10} }} \\ \\ \tt{\purple{\implies a = 19 \: {m/s}^{2} }} \\ \\ \bold{\pink{Again:}} \\ \\ \tt{\green{\implies F_{1} - F_{2} - \mu N= ma}} \\ \\ \tt{\red{\implies 200 - 50 - 0.1 \times 150 = 10 \times a}} \\ \\ \tt{\orange{\implies 135 = 10 \times a}} \\ \\ \tt{\pink{\implies a = \frac{135}{10}}} \\ \\ \tt {\purple{\implies a = 13.5\: {m/s}^{2}} }$