Question

A burglar attempts to drag a 108 kg metal safe across a polished wood floor Assume that the coefficient of static friction is 0.4, the coefficient of kinetic friction is 0.3, and that the burglar can apply a pushing force of 534 N on the metal safe. What is the acceleration of the metal safe across floor?

Answer 1

Answer:

2.00 m/s²

Explanation:

Given

The Mass of the metal safe, M = 108kg

Pushing force applied by the burglar,  F = 534 N

Co-efficient of kinetic friction, $\mu_k$ = 0.3

Now,

The force against the kinetic friction is given as:

$f = \mu_k N = u_k Mg$

Where,

N = Normal reaction

g= acceleration due to the gravity

Substituting the values in the above equation, we get

$f = 0.3\times108\times9.8$

or

$f = 317.52N$

Now, the net force on to the metal safe is

$F_{Net}= F-f$

Substituting the values in the equation we get

 $F_{Net}= 534N-317.52N$

or

$F_{Net}= 216.48$

also,

 

$F_{Net}= M\times$acceleration of the safe

Therefore, the acceleration of the metal safe will be

acceleration of the safe=$\frac{F_{Net}}{M}$

or

 acceleration of the safe=$\frac{216.48}{108}$

or

 

acceleration of the safe=$2.00 m/s^2$

Hence, the acceleration of the metal safe will be  2.00 m/s²