A block of mass $1 kg$ lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is $0.6$ . If the acceleration of the truck is $5 {m/s}^{2}$ , the frictional force acting on the block in newtons is: (take $g = 10 {ms}^{- 2}$ )

Open in App

Solution

Maximum friction $= μ m g = 0.6 \times 10 \times 1 = 6 N$
Pseudo force $= m a = 5 N$
so required frictional force is only $5 N$ , although maximum frictional force available is $6 N$ .

So, the frictional force acting on the block is $5 N$

Suggest Corrections

Similar questions

Q. A block of mass

1 kg

lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is

0.6

. If the acceleration of the truck is

5 {m/s}^{2}

, the frictional force acting on the block is,
[Take

g = 10 {m/s}^{2}

]

Q. Fill in the blanks.
A block of mass

1 k g

lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is

0.6.

If the acceleration of the truck is

5 m / s^{2},

the frictional force acting on the block is ........ Newtons.

Q. A block of mass

1 k g

lies on a horizontal surface in a truck, the coefficient of static friction between the block and the surface is

0.6

. What is the force of friction on the block if the acceleration of the truck is

5 m / s^{2}

?
(Take

g = 10 m / s^{2}

)

A block of mass $1 k g$ lies on horizontal surface in a truck. The coefficient of static friction between the block and the surface is $μ_{s} = 0.6$ . If the acceleration of the truck is $5 m / s^{2}$ , the friction force acting on the block is

Q.
A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is

5 m s^{- 2}

, the frictional force acting on the block is ...............N.

Join BYJU'S Learning Program

A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is 0.6. If the acceleration of the truck is 5 m/s2, the frictional force acting on the block in newtons is: (take g=10 ms−2)

Maximum friction =μmg=0.6×10×1=6 N Pseudo force =ma=5 N so required frictional force is only 5 N, although maximum frictional force available is 6 N. So, the frictional force acting on the block is 5 N

A block of mass $1 kg$ lies on a horizontal surface in a truck. The coefficient of static friction between the block and the surface is $0.6$ . If the acceleration of the truck is $5 {m/s}^{2}$ , the frictional force acting on the block in newtons is: (take $g = 10 {ms}^{- 2}$ )

Maximum friction $= μ m g = 0.6 \times 10 \times 1 = 6 N$
Pseudo force $= m a = 5 N$
so required frictional force is only $5 N$ , although maximum frictional force available is $6 N$ .

So, the frictional force acting on the block is $5 N$