## Physics (10th Edition)

The speed of the object at $s=20m$ is $7.07m/s$
The net work done by the force is the area of the red shape as portrayed in the force-displacement graph. $$\sum W=\frac{1}{2}(10m+20m)\times10N=150J$$ The work-energy theorem states that $$\sum W=\frac{1}{2}m(v_f^2-v_0^2)$$ We know $v_0=0$ and $m=6kg$. Therefore, $$3v_f^2=150$$ $$v_f=7.07m/s$$ which is the final speed of the object at $s=20m$
{"ft_lang_label":"__label__en","ft_lang_prob":0.83576703,"math_prob":1.0000019,"size":418,"snap":"2023-14-2023-23","text_gpt3_token_len":155,"char_repetition_ratio":0.10869565,"word_repetition_ratio":0.07017544,"special_character_ratio":0.39473686,"punctuation_ratio":0.05319149,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":1.0000092,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-07T13:10:21Z\",\"WARC-Record-ID\":\"<urn:uuid:8eaf0ad6-56ce-4d2b-95cd-e4c662a13cb2>\",\"Content-Length\":\"54134\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ac9e1e34-5fc5-429e-a34f-06e320d3f2d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:d42a0608-17fb-4b52-a899-b94da1de4ffe>\",\"WARC-IP-Address\":\"3.215.61.136\",\"WARC-Target-URI\":\"https://www.gradesaver.com/textbooks/science/physics/physics-10th-edition/chapter-6-work-and-energy-problems-page-170/75\",\"WARC-Payload-Digest\":\"sha1:T5IFT2LPT2MFP4DRECK3VLA6OKL76MBW\",\"WARC-Block-Digest\":\"sha1:5CJ2BF47KFKSP3YDUASGRMD77UEC4HQ2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653764.55_warc_CC-MAIN-20230607111017-20230607141017-00379.warc.gz\"}"}