The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open and The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s⁻². At what distance from the starting point does the box fall of the truck? (Ignore the size of the box).
Answers
- Rear side of the truck = 40 kg of mass coefficient friction between Box and surface .
- Acceleration = 2m/s -2
- Fall of the truck ignore the size of The box = ???
- Main things Given
- Box mass = 40 kg
- U = 0.15
- Speed = 5m
- Acceleration = 2m/s -2
- Force = 40 × 2 = 80n
So, the truck Acceleration is = umg = Force
- = 0.15 × 40 × 10 = 60 .
- Net force is block
- For Net force = 80 - 60 = 20
So, = Force net / mass = 20 / 40
- = 0.5 m/s
Therefore,
Distance = the truck had travelled
Answer:
Given:
Rear side of the truck = 40 kg of mass coefficient friction between Box and surface .
{\small{\bf{{\underline{\pink{ = 0:15}}}}}}
=0:15
Acceleration = 2m/s -2
{\large{\bf{\underline{\underline{\red{→To \: Find \: }}}}}}
→ToFind
Fall of the truck ignore the size of The box = ???
{\large{\bf{\underline{\underline{\red{→According \: to \: the \: question }}}}}}
→Accordingtothequestion
Main things Given
Box mass = 40 kg
U = 0.15
Speed = 5m
Acceleration = 2m/s -2
{\large{\bf{\underline{\underline{\red{→Force = Mass × Acceleration }}}}}}
→Force=Mass×Acceleration
Force = 40 × 2 = 80n
So, the truck Acceleration is = umg = Force
= 0.15 × 40 × 10 = 60 .
Net force is block
For Net force = 80 - 60 = 20
So, = Force net / mass = 20 / 40
= 0.5 m/s
{\small{\bf{\underline{\underline{\red{→speed \: = \:0 + \frac{1}{2} \times 0.5 \times t}}}}}}
→speed=0+
2
1
×0.5×t
Therefore,
Distance = the truck had travelled
{\large{\bf{\underline{\underline{\blue{→ t = \sqrt{20} \: speed }}}}}}
→t=
20
speed