Question

A burglar attempts to drag a 110 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 524 N on the metal safe. What is the acceleration of the metal safe across floor?

Answer 1

Answer:

a = 1.82 m/s/s

Explanation:

Here give that

static friction coefficient on the floor = 0.4

kinetic friction coefficient on the floor = 0.3

mass of the box = 110 kg

now the normal force on the box due to ground is given as

$F_n = mg$

$F_n = (110)(9.81)$

$F_n = 1079.1 N$

Now the minimum force required to move the box on the rough surface in horizontal direction is given as

$F = \mu_s F_n$

$F = 0.4(1079.1) = 431.64 N$

now it is given that force applied by the burglar is 524 N which is more than the static friction

so here the box will have kinetic friction on it

$F_k = \mu_k F_n$

$F_k = (0.3)(1079.1) = 323.7 N$

now the net force on the box is given as

$F_{net} = 524 - 323.7$

$F_{net} = 200.3 N$

now the acceleration of the box is given by Newton's II law

$a = \frac{F_{net}}{m}$

$a = \frac{200.3}{110} = 1.82 m/s^2$

Answer 2

The acceleration of the metal safe across floor is 1.83 m/s².

The given parameters;

• mass of the metal, m = 110 kg
• applied horizontal force, Fₓ = 524 N
• coefficient kinetic friction, μk = 0.3

The weight of the metal safe is calculated as;

Fₙ = mg

Fₙ = 110 x 9.8

Fₙ = 1078 N

The frictional force on metal safe is calculated as follows;

$F_k = \mu_k F_n\\\\F_k = 0.3\times 1078\\\\F_k = 323.4 \ N$

The net horizontal force on the metal safe is calculated as follows;

$F_{net} = F_x - F_k\\\\F_{net} = 524 - 323.4\\\\F_{net} = 200.6 \ N$

The acceleration of the metal safe across floor is calculated as follows;

F = ma

$a = \frac{F_{net}}{m} \\\\a = \frac{200.6}{110} \\\\a = 1.83 \ m/s^2$

Thus, the acceleration of the metal safe across floor is 1.83 m/s².

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