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Komok [63]
3 years ago
14

If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o

n dry concrete where the coefficient of kinetic friction is 0.7 and the coefficient of static friction is 1.
Physics
1 answer:
Scilla [17]3 years ago
8 0

Answer:

Maximum acceleration will be equal to 3.43m/sec^2

Explanation:

We have given coefficient of kinetic friction \mu _k=0.7

And coefficient of static friction \mu _s=1

Acceleration due to gravity g=9.8m/sec^2

When truck moves maximum force will be equal to F=\mu _kmg

It is given that half of the weight is supported by its drive wheels

So force required =\frac{\mu _kmg}{2}

From newtons law maximum acceleration will be equal to a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2

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