Question

A truck starts from rest and rolls downs a hill with constant acceleration it travelles distance of 500 m in 10 sec. find it's acceleration . find the force acting on it if it's mass is 6000 kg​

Answer 1

Answer:

Solution

initial velocity u = 0

final velocity v = distance/ time

v = 500 m / 10 sec

v = 50 m/s

v = u + at

50 m/s = 0 + a×10 sec

a = 50 m/s / 10 s

a = 5 m/s^2

Force = mass × acceleration

F = m×a

F = 6000 kg × 5 m/s^2

F = 30,000 kgm/s^2

F = 30,000 N

Hope this solves your question

Answer 2

Given : Initial velocity (u) = 0m/s

$~~~~~~~~~~~~$Final velocity (v) =${ \sf{ \dfrac{500}{10} = 50ms}}$

$~~~~~~~~~~~~$Time = 10 seconds

To find : Find acceleration & Force.

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$\underline{ \frak{as~we~know~that~ : }}$

• $\underset{ \blue{ \sf \: acceleration \ formula }}{ \underbrace{ \boxed{ \frak{ \pink{a~ = ~ \dfrac{v~ - ~u}{t}}}}}}$

$~$

$\qquad \qquad{ \sf : \implies{a~ = ~ \dfrac{50 - 0}{10}}}$

$\qquad \qquad { \sf : \implies{a~ = ~ \dfrac{50}{10}}}$

$\qquad \qquad : \implies{ \underline{ \boxed{ \frak{ \purple{ a~ = ~ \: 50m {s}^{2} }}}}} \bigstar$

$~$

$\underline{ \frak{ as~we~know~that~ : }}$

• $\underset{ \blue{ \sf force \ \: formula}}{ \underbrace{ \boxed{ \frak{ \pink{ f ~ = ~ma}}}}}$

$~$

$\qquad \qquad{ \sf : \implies{f ~ = ~6000 ~\times~ 5}}$

$\qquad \qquad : \implies{ \underline{ \boxed{ \frak{ \pink{ f~ = ~30000 }}}}} \star$

$~$

Hence,

$\therefore \underline{ \sf{ the~force~acting~on~truck~is~ \bf{ \underline{30000n}}}}$

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$\star{ \underline{\boxed{ \frak{ \pink{MagicalWworld07}}}}} \star$