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A parcel of mass 10 kg rests on a lorry.When the lorry accelerates at 1.5 m/s^2,the parcel is just about to slide backwards. What is the co-efficient of friction between the parcel and the lorry?​

User Cybaek
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2 Answers

2 votes

Final answer:

To find the coefficient of friction (μs) between the parcel and the lorry, you divide the truck's acceleration (1.5 m/s²) by the acceleration due to gravity (9.8 m/s²), resulting in an approximate value of 0.153.

Step-by-step explanation:

To find the coefficient of friction between the parcel and the lorry, we need to consider the forces acting on the parcel when the lorry accelerates. The force of static friction (ℒs) is what keeps the parcel from sliding. This force can be calculated using the formula ℒs = μs× N, where μs is the coefficient of static friction, and N is the normal force. In this case, the normal force is equal to the weight of the parcel, which is mass (m) times the acceleration due to gravity (g), so N = m × g. The static frictional force must be equal to the product of the parcel's mass and the truck's acceleration to prevent the parcel from sliding (ℒs = m × a). Putting these equations together, we can derive μs by dividing the parcel's mass times the truck's acceleration with the normal force: μs = (m × a) / (m × g), which simplifies to μs = a / g.

Given that the parcel's mass is 10 kg, the acceleration due to gravity is 9.8 m/s², and the truck's acceleration is 1.5 m/s², the coefficient of friction is calculated as follows:

μs = a / g = 1.5 m/s² / 9.8 m/s².

Performing the calculation gives us:μs ≈ 0.153

User Barbaros Alp
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1 vote

Answer:

The coefficient of static friction, μ = 0.153

Step-by-step explanation:

Given data,

The mass of the parcel, m = 20 kg

The acceleration of the lorry, a = 1.5 m/s²

The normal force acting on the parcel,

f = m x g

= 20 x 9.8

= 196 N

The static frictional force acting on the lorry,

F = 20 x 1.5

= 30 N

The coefficient of static is

μ = F / f

= 30 / 196

= 0.153

Hence, the coefficient of static friction, μ = 0.153

User Mahani
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