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How much work must Denise do to drag her basket of laundry of mass $5.0 \mathrm{kg}$ a distance of $5.0 \mathrm{m}$ along a floor, if the force she exerts is a constant $30.0 \mathrm{N}$ at an angle of $60.0^{\circ}$ with the horizontal?

$$W=75 \mathrm{J}$$

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Simon Fraser University

Hope College

University of Sheffield

University of Winnipeg

So this question is asking us to calculate the work done on a basket of laundry if it has moved a distance of five meters and the force that is exerted on the basket. Is that an angle? So let's start by quickly capturing all the information that is given to us in the problem here. I've already drawn the laundry basket, but let's at in the five meters that the laundry basket goes and let's add in the Applied Force. So we're told that the force is 30 Newton's and it's happening at an angle of 60 degrees with the horizontal. So this is all the information. One other piece of information. The mass of the laundry basket is five kilograms. So to solve a problem like this were asked to solve for the work. In order to solve a problem like this, we need a formula that relates the information that we have in this case force distance and mass and an angle if that's relevant to. So the formula that we're gonna use here is work equals force. Times Delta are sometimes called distance, sometimes called displacement times Coast data. So this is the scaler non vector version of the formula. Another version of this is the work equals forces of Vector dotted with displacement. Um, but because the information we're given is an angle and, um, directional form, we can use just the scaler version anyways, because that's the version that we would use if we were given them in vector notation. I'd probably used the other one. So plugging in the numbers that we have, we have a force that is 30 Newton's. We have a displacement that is five meters, and we're gonna take co sine of the angle between them, which in this case, is 60 degrees punching metal into the calculator. We got 30 times five, which 150 coastline of 60 is 1/2 or a 600.5, and altogether this is going to give me 75 jewels work

University of Winnipeg