Question

7- find apparent weight of a 220.0 lbs (pounds) person when elevator goes up with acceleration of 1.40 m/s². (b) when elevator goes down with acceleration of 1.40 m/s². (c) when elevator goes with constant velocity up? find your final answers for apparent weight by pounds.(1lb= 4.44newton)

Explanation:

Step1: Convert the mass from pounds to Newtons

The actual weight \(W = 220.0\times4.44\space N\).

$$W=220\times4.44 = 976.8\space N$$

The mass \(m=\frac{W}{g}\), where \(g = 9.8\space m/s^{2}\).

$$m=\frac{976.8}{9.8}=99.67\space kg$$

Step2: (a) Elevator going up

Using Newton's second law \(F_{N}-mg=ma\) (where \(F_{N}\) is the normal force, which is the apparent weight).

$$F_{N}=m(g + a)$$

Substitute \(m = 99.67\space kg\), \(g=9.8\space m/s^{2}\), \(a = 1.40\space m/s^{2}\)

$$F_{N}=99.67\times(9.8 + 1.40)=99.67\times11.2$$

$$F_{N}=1116.3\space N$$

Convert back to pounds: \(\frac{1116.3}{4.44}\approx251.4\space lbs\)

Step3: (b) Elevator going down

Using \(mg - F_{N}=ma\) (since the acceleration is in the downward direction)

$$F_{N}=m(g - a)$$

Substitute \(m = 99.67\space kg\), \(g = 9.8\space m/s^{2}\), \(a=1.40\space m/s^{2}\)

$$F_{N}=99.67\times(9.8- 1.40)=99.67\times8.4$$

$$F_{N}=837.2\space N$$

Convert back to pounds: \(\frac{837.2}{4.44}\approx188.6\space lbs\)

Step4: (c) Elevator moving with constant velocity

When \(a = 0\), using \(F_{N}-mg=0\) (from \(F_{N}-mg=ma\) with \(a = 0\))

$$F_{N}=mg$$

Since \(W=mg = 976.8\space N\)
Convert back to pounds: \(\frac{976.8}{4.44}=220.0\space lbs\)

Answer:

(a) \(251.4\space lbs\)
(b) \(188.6\space lbs\)
(c) \(220.0\space lbs\)